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Exercise Set 1

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Exercises

Exercise 1
Question

Use mathematical induction to prove:

12+22+32+⋯+n2=n(n+1)(2n+1)61^{2}+2^{2}+3^{2}+\dots+n^{2} = \frac{n(n+1)(2n+1)}{6}

for any positive integer nn.

Show solution ↓
Solution

Step 1: Base Case

For n=1n=1:

12=11^{2}=11(1+1)(2⋅1+1)6=1⋅2⋅36=1\frac{1(1+1)(2\cdot1+1)}{6} = \frac{1\cdot2\cdot3}{6} = 1

So the formula is true for n=1n=1.

Step 2: Inductive Hypothesis

Assume the formula holds for n=kn=k:

12+22+⋯+k2=k(k+1)(2k+1)6.1^{2}+2^{2}+\dots+k^{2} = \frac{k(k+1)(2k+1)}{6}.

Step 3: Inductive Step

For n=k+1n=k+1:

12+22+⋯+k2+(k+1)2.1^{2}+2^{2}+\dots+k^{2}+(k+1)^{2}.

Using the hypothesis:

=k(k+1)(2k+1)6+(k+1)2.= \frac{k(k+1)(2k+1)}{6}+(k+1)^{2}.

Factor (k+1)(k+1):

=(k+1)(k(2k+1)6+(k+1)).= (k+1)\left(\frac{k(2k+1)}{6}+(k+1)\right).

Combine into one fraction:

=(k+1)(k(2k+1)+6(k+1))6.= \frac{(k+1)\left(k(2k+1)+6(k+1)\right)}{6}.

Simplify numerator:

k(2k+1)+6(k+1)=2k2+7k+6=(2k+3)(k+2).k(2k+1)+6(k+1) = 2k^{2}+7k+6 = (2k+3)(k+2).

Thus,

=(k+1)(k+2)(2k+3)6.= \frac{(k+1)(k+2)(2k+3)}{6}.

Since 2k+3=2(k+1)+12k+3=2(k+1)+1, we obtain

(k+1)((k+1)+1)(2(k+1)+1)6.\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}.

This matches the formula for n=k+1n=k+1.

Conclusion

By mathematical induction, the formula holds for all positive integers nn.

Exercise 2
Question

Use mathematical induction to prove:

3n−1 is divisible by 23^{n}-1 \text{ is divisible by } 2

for any positive integer nn.

Show solution ↓
Solution

Base Case: For n=1n=1,

31−1=2,3^{1}-1=2,

which is divisible by 2.

Inductive Hypothesis: Assume 3k−13^{k}-1 is divisible by 2.
Then 3k−1=2m3^{k}-1=2m for some integer mm.

Inductive Step:

3k+1−1=3⋅3k−1=3(3k−1)+2.3^{k+1}-1=3\cdot3^{k}-1 =3(3^{k}-1)+2.

Since 3k−1=2m3^{k}-1=2m,

=3(2m)+2=6m+2=2(3m+1).=3(2m)+2=6m+2=2(3m+1).

Thus divisible by 2.

Therefore, true for all nn.

Exercise 3
Question

Use mathematical induction to prove:

a+ar+ar2+⋯+arn−1=a(1−rn)1−r,r≠1.a+ar+ar^{2}+\dots+ar^{n-1} = \frac{a(1-r^{n})}{1-r}, \quad r\ne1.
Show solution ↓
Solution

Base Case: For n=1n=1,

LHS=a,RHS=a(1−r)1−r=a.\text{LHS}=a, \quad \text{RHS}=\frac{a(1-r)}{1-r}=a.

Inductive Hypothesis: Assume true for n=kn=k:

a+ar+⋯+ark−1=a(1−rk)1−r.a+ar+\dots+ar^{k-1} = \frac{a(1-r^{k})}{1-r}.

Inductive Step: Add arkar^{k}:

=a(1−rk)1−r+ark.= \frac{a(1-r^{k})}{1-r}+ar^{k}.

Write over common denominator:

=a(1−rk)+ark(1−r)1−r.= \frac{a(1-r^{k})+ar^{k}(1-r)}{1-r}.

Simplify numerator:

a−ark+ark−ark+1=a(1−rk+1).a - ar^{k} + ar^{k} - ar^{k+1} = a(1-r^{k+1}).

Thus,

=a(1−rk+1)1−r.= \frac{a(1-r^{k+1})}{1-r}.

True for all nn.

Exercise 4
Question

Use mathematical induction to prove:

2n≥n+12^{n}\ge n+1

for any positive integer nn.

Show solution ↓
Solution

Base Case: For n=1n=1,

21=2≥2.2^{1}=2 \ge 2.

Inductive Hypothesis: Assume 2k≥k+12^{k}\ge k+1.

Inductive Step:

2k+1=2⋅2k≥2(k+1)=2k+2.2^{k+1}=2\cdot2^{k} \ge2(k+1) =2k+2.

Since 2k+2≥k+22k+2 \ge k+2 for k≥1k\ge1,

2k+1≥k+2.2^{k+1}\ge k+2.

Thus true for all nn.

Exercise 5
Question

Write the first four terms of the sequence:

{(−1)n2nn!}n=1∞.\left\{(-1)^{n}\frac{2^{n}}{n!}\right\}_{n=1}^{\infty}.
Show solution ↓
Solution

For n=1n=1:

(−1)21=−2.(-1)\frac{2}{1}= -2.

n=2n=2:

(+1)42=2.(+1)\frac{4}{2}=2.

n=3n=3:

(−1)86=−43.(-1)\frac{8}{6}= -\frac{4}{3}.

n=4n=4:

(+1)1624=23.(+1)\frac{16}{24}= \frac{2}{3}.

First four terms:

−2,  2,  −43,  23.-2,\;2,\;-\frac{4}{3},\;\frac{2}{3}.
Exercise 6
Question

Write the first four terms of the sequence:

{sin⁡(nπ2)−cos⁡(nπ)}n=0∞.\left\{\sin\left(\frac{n\pi}{2}\right)-\cos(n\pi)\right\}_{n=0}^{\infty}.
Show solution ↓
Solution

n=0n=0:

0−1=−1.0-1=-1.

n=1n=1:

1−(−1)=2.1-(-1)=2.

n=2n=2:

0−1=−1.0-1=-1.

n=3n=3:

−1−(−1)=0.-1-(-1)=0.

First four terms:

−1,  2,  −1,  0.-1,\;2,\;-1,\;0.
Exercise 7
Question

Determine whether the sequence

{n−1n}n=1∞\left\{\frac{n-1}{n}\right\}_{n=1}^{\infty}

is convergent or divergent. Find its limit if convergent.

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Solution
n−1n=1−1n.\frac{n-1}{n}=1-\frac{1}{n}.

As n→∞n\to\infty, 1n→0\frac{1}{n}\to0.

Thus,

lim⁡n→∞n−1n=1.\lim_{n\to\infty}\frac{n-1}{n}=1.

The sequence converges to 1.

Exercise 8
Question

Determine whether the sequence

{n2n2+1n2}n=1∞\left\{\frac{\sqrt{\frac{n^2}{n^2 +1}}}{\sqrt{n^2}}\right\}_{n=1}^{\infty}

is convergent or divergent. Find its limit if convergent.

Show solution ↓
Solution

Simplify:

n2=n,n2n2+1=nn2+1.\sqrt{n^2}=n, \quad \sqrt{\frac{n^2}{n^2+1}} =\frac{n}{\sqrt{n^2+1}}.

Thus,

nn2+1⋅1n=11+1n2.\frac{n}{\sqrt{n^2+1}}\cdot\frac{1}{n} = \frac{1}{\sqrt{1+\frac{1}{n^2}}}.

As n→∞n\to\infty,

11+0=1.\frac{1}{\sqrt{1+0}}=1.

The sequence converges to 1.

Exercise 9
Question

Determine whether the sequence

{(n+1n+2)n+3}n=1∞\left\{\left(\frac{n+1}{n+2}\right)^{n+3}\right\}_{n=1}^{\infty}

is convergent or divergent. Find its limit if convergent.

Show solution ↓
Solution

Rewrite:

n+1n+2=1−1n+2.\frac{n+1}{n+2} = 1-\frac{1}{n+2}.

Thus,

(1−1n+2)n+3.\left(1-\frac{1}{n+2}\right)^{n+3}.

Let m=n+2m=n+2. Then expression becomes

(1−1m)m+1.\left(1-\frac{1}{m}\right)^{m+1}.

We know

(1−1m)m→1e.\left(1-\frac{1}{m}\right)^m \to \frac{1}{e}.

Therefore,

(1−1m)m+1=(1−1m)m(1−1m)→1e.\left(1-\frac{1}{m}\right)^{m+1} = \left(1-\frac{1}{m}\right)^m \left(1-\frac{1}{m}\right) \to \frac{1}{e}.

Hence the sequence converges to 1e\frac{1}{e}.

Exercise 10
Question

Determine whether the sequence

{2n+2n2}n=1∞\left\{\frac{2^{n}+2}{n^{2}}\right\}_{n=1}^{\infty}

is convergent or divergent.

Show solution ↓
Solution

As n→∞n\to\infty, 2n2^n grows much faster than n2n^2.

Thus,

2n+2n2∼2nn2.\frac{2^n+2}{n^2} \sim \frac{2^n}{n^2}.

Since exponential growth dominates polynomial growth, the terms grow without bound.

Therefore the sequence diverges to ∞\infty.

Exercise 11
Question

Determine whether the sequence

{4+cos⁡nn}n=1∞\left\{4+\frac{\cos n}{\sqrt n}\right\}_{n=1}^{\infty}

is convergent or divergent. Find its limit if convergent.

Show solution ↓
Solution

We know

−1≤cos⁡n≤1.-1\le \cos n \le 1.

Thus

−1n≤cos⁡nn≤1n.-\frac{1}{\sqrt n}\le \frac{\cos n}{\sqrt n} \le \frac{1}{\sqrt n}.

Since 1n→0\frac{1}{\sqrt n}\to0,

cos⁡nn→0.\frac{\cos n}{\sqrt n}\to0.

Therefore,

4+cos⁡nn→4.4+\frac{\cos n}{\sqrt n}\to4.

The sequence converges to 4.

Exercise 12
Question

Determine whether the sequence

{n−n2}n=1∞\{n-n^{2}\}_{n=1}^{\infty}

is monotonic.

Show solution ↓
Solution
an=n−n2=−n2+n.a_n=n-n^2=-n^2+n.

Compute difference:

an+1−an=((n+1)−(n+1)2)−(n−n2).a_{n+1}-a_n = ((n+1)-(n+1)^2)-(n-n^2).

Simplify:

=−2n.= -2n.

Since -2n<0 for n≥1n\ge1, the sequence is decreasing.

Thus it is monotonic decreasing.

Exercise 13
Question

Determine whether the sequence

{2n(n+1)!}n=1∞\left\{\frac{2^{n}}{(n+1)!}\right\}_{n=1}^{\infty}

is monotonic.

Show solution ↓
Solution

Consider ratio:

an+1an=2n+1/(n+2)!2n/(n+1)!=2n+2.\frac{a_{n+1}}{a_n} = \frac{2^{n+1}/(n+2)!}{2^n/(n+1)!} = \frac{2}{n+2}.

For n≥1n\ge1,

\frac{2}{n+2}<1.

Thus a_{n+1}<a_n.

The sequence is decreasing.

Exercise 14
Question

Determine whether the sequence

{nn+1}n=1∞\left\{\frac{n}{n+1}\right\}_{n=1}^{\infty}

is monotonic.

Show solution ↓
Solution
an=nn+1.a_n=\frac{n}{n+1}.

Compute difference:

an+1−an=n+1n+2−nn+1.a_{n+1}-a_n = \frac{n+1}{n+2}-\frac{n}{n+1}.

Combine:

=\frac{(n+1)^2-n(n+2)}{(n+2)(n+1)} = \frac{1}{(n+2)(n+1)}>0.

Thus increasing.

The sequence is monotonic increasing.

Exercise 15
Question

Determine whether the series

∑n=1∞(−3)n22n\sum_{n=1}^{\infty}\frac{(-3)^{n}}{2^{2n}}

is convergent or divergent. Find its sum if convergent.

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Solution
22n=4n.2^{2n}=4^n.

Thus series becomes

∑(−34)n.\sum \left(\frac{-3}{4}\right)^n.

This is geometric with ratio r=−34r=-\frac34.

Since |r|<1, it converges.

Sum:

r1−r=−341+34=−37.\frac{r}{1-r} = \frac{-\frac34}{1+\frac34} = -\frac{3}{7}.
Exercise 16
Question

Determine whether the series

∑n=1∞32n(n+1)\sum_{n=1}^{\infty}\frac{32}{n(n+1)}

is convergent or divergent. Find its sum if convergent.

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Solution

Use partial fractions:

32n(n+1)=32(1n−1n+1).\frac{32}{n(n+1)} = 32\left(\frac1n-\frac1{n+1}\right).

This telescopes.

Sum becomes

32(1−lim⁡n→∞1n+1)=32.32\left(1-\lim_{n\to\infty}\frac1{n+1}\right) =32.

Thus convergent with sum 3232.

Exercise 17
Question

Determine whether the series

∑n=1∞23(2n−1)31−n\sum_{n=1}^{\infty}\frac{23(2^{n-1})}{3^{1-n}}

is convergent or divergent. Find its sum if convergent.

Show solution ↓
Solution

First simplify the term.

31−n=33n.3^{1-n}=\frac{3}{3^n}.

Thus,

23(2n−1)31−n=23(2n−1)3n3=2332n−13n.\frac{23(2^{n-1})}{3^{1-n}} = 23(2^{n-1})\frac{3^n}{3} = \frac{23}{3}2^{n-1}3^n.=233⋅2n2⋅3n=236(6n).= \frac{23}{3}\cdot \frac{2^n}{2}\cdot 3^n = \frac{23}{6}(6^n).

Thus the series becomes

∑2366n.\sum \frac{23}{6}6^n.

Since 6n6^n grows without bound, the terms do not approach 0.

Therefore the series diverges.

Exercise 18
Question

Determine whether the series

∑n=3∞3n2−2n\sum_{n=3}^{\infty}\frac{3}{n^{2}-2n}

is convergent or divergent. Find its sum if convergent.

Show solution ↓
Solution

Factor denominator:

n2−2n=n(n−2).n^2-2n=n(n-2).

Use partial fractions:

3n(n−2)=An+Bn−2.\frac{3}{n(n-2)} = \frac{A}{n}+\frac{B}{n-2}.

Solving gives:

3n(n−2)=3/2n−2−3/2n.\frac{3}{n(n-2)} = \frac{3/2}{n-2}-\frac{3/2}{n}.

Thus series telescopes:

32(11+12−lim⁡n→∞1n−1n−1).\frac{3}{2}\left(\frac1{1}+\frac1{2}-\lim_{n\to\infty}\frac1n-\frac1{n-1}\right).

Remaining terms:

32(1+12)=32⋅32=94.\frac{3}{2}\left(1+\frac12\right) = \frac{3}{2}\cdot\frac32 = \frac{9}{4}.

Convergent with sum 94\frac{9}{4}.

Exercise 19
Question

Determine whether the series

∑n=1∞n3nsin⁡(2nπ)\sum_{n=1}^{\infty}n^{3n}\sin(2n\pi)

is convergent or divergent.

Show solution ↓
Solution

We know

sin⁡(2nπ)=0\sin(2n\pi)=0

for all integers nn.

Thus every term is zero.

So the series equals 0.

Therefore it converges (sum = 0).

Exercise 20
Question

Find the value of aa which makes the series

1−(1+a)+(1+a)2−(1+a)3+…1-(1+a)+(1+a)^{2}-(1+a)^{3}+\dots

convergent.

Show solution ↓
Solution

This is a geometric series:

∑(−1)n(1+a)n.\sum (-1)^n(1+a)^n.

Common ratio:

r=−(1+a).r=-(1+a).

Geometric series converges if

|r|<1.

Thus

|1+a|<1.

Solve:

-1<1+a<1.-2<a<0.
Exercise 21
Question

Use the integral test to show that

∑n=1∞1np\sum_{n=1}^{\infty}\frac{1}{n^{p}}

is convergent when p>1.

Show solution ↓
Solution

Consider

∫1∞1xpdx.\int_1^\infty \frac1{x^p}dx.=∫1∞x−pdx=[x1−p1−p]1∞.= \int_1^\infty x^{-p}dx = \left[\frac{x^{1-p}}{1-p}\right]_1^\infty.

If p>1, then 1-p<0.

Thus x1−p→0x^{1-p}\to0 as x→∞x\to\infty.

So integral is finite.

Therefore the series converges for p>1.

Exercise 22
Question

Determine whether the series

∑n=1∞1n+2\sum_{n=1}^{\infty}\frac{1}{n+2}

is convergent or divergent.

Show solution ↓
Solution
1n+2∼1n.\frac{1}{n+2}\sim\frac1n.

Since harmonic series diverges,

∑1n+2\sum \frac1{n+2}

also diverges by comparison.

Exercise 23
Question

Determine whether the series

∑n=1∞12n+n2\sum_{n=1}^{\infty}\frac{1}{2^{n}+n^{2}}

is convergent or divergent.

Show solution ↓
Solution

Since

2^n+n^2>2^n,\frac1{2^n+n^2}<\frac1{2^n}.

Geometric series ∑12n\sum \frac1{2^n} converges.

Thus by comparison, given series converges.

Exercise 24
Question

Determine whether the series

∑n=1∞(n+2)!222n\sum_{n=1}^{\infty}\frac{(n+2)!}{222^{n}}

is convergent or divergent.

Show solution ↓
Solution

Use Ratio Test.

an+1an=(n+3)!/222n+1(n+2)!/222n=n+3222.\frac{a_{n+1}}{a_n} = \frac{(n+3)!/222^{n+1}}{(n+2)!/222^n} = \frac{n+3}{222}.

As n→∞n\to\infty, ratio →∞\to\infty.

Thus series diverges.

Exercise 25
Question

Determine whether the series

∑n=1∞(2n)!(n+2)!\sum_{n=1}^{\infty}\frac{(2n)!}{(n+2)!}

is convergent or divergent.

Show solution ↓
Solution

Factorial growth in numerator is much faster than denominator.

Terms grow rapidly and do not approach zero.

Therefore the series diverges.

Exercise 26
Question

Determine whether the series

∑n=1∞ln⁡(n3)3n\sum_{n=1}^{\infty}\frac{\sqrt{\ln(n^{3})}}{3n}

is convergent or divergent.

Show solution ↓
Solution
ln⁡(n3)=3ln⁡n.\ln(n^3)=3\ln n.

Thus term behaves like

ln⁡nn.\frac{\sqrt{\ln n}}{n}.

Since

\frac{\sqrt{\ln n}}{n} > \frac1n \text{ eventually},

and harmonic diverges,

the series diverges.

Exercise 27
Question

Determine whether the series

∑n=2∞n(n3+3)(3n−3)\sum_{n=2}^{\infty}\frac{n}{(n^{3}+3)(3n-3)}

is convergent or divergent.

Show solution ↓
Solution

For large nn,

(n3+3)(3n−3)∼3n4.(n^{3}+3)(3n-3)\sim 3n^4.

Thus

n(n3+3)(3n−3)∼n3n4=13n3.\frac{n}{(n^{3}+3)(3n-3)} \sim \frac{n}{3n^4} = \frac{1}{3n^3}.

Since ∑1n3\sum \frac1{n^3} converges (p=3>1),
by comparison the series converges.

Exercise 28
Question

Determine whether the series

∑n=3∞13−n(n−3)!\sum_{n=3}^{\infty}\frac{1}{3^{-n}(n-3)!}

is convergent or divergent.

Show solution ↓
Solution
13−n(n−3)!=3n(n−3)!.\frac{1}{3^{-n}(n-3)!} = \frac{3^n}{(n-3)!}.

Factorial grows slower than 3n3^n here in denominator shift form.

Using Ratio Test:

an+1an=3n+1/(n−2)!3n/(n−3)!=3n−2.\frac{a_{n+1}}{a_n} = \frac{3^{n+1}/(n-2)!}{3^n/(n-3)!} = \frac{3}{n-2}.

As n→∞n\to\infty, ratio →0\to0.

Thus series converges absolutely.

Exercise 29
Question

Determine whether the series

∑n=1∞2⋅33n+3n3−33n+1\sum_{n=1}^{\infty}\frac{2\cdot3^{3n}+3^{n}}{3-3^{3n+1}}

is convergent or divergent.

Show solution ↓
Solution

Dominant term is 33n3^{3n}.

Expression behaves like

33n−33n+1=−13.\frac{3^{3n}}{-3^{3n+1}} = -\frac{1}{3}.

Terms do not approach 0.

Thus the series diverges.

Exercise 30
Question

Determine whether the series

∑n=1∞cos⁡(1+n+2n1+n+n2)\sum_{n=1}^{\infty}\cos\left(\frac{1+n+2n}{1+n+n^{2}}\right)

is convergent or divergent.

Show solution ↓
Solution

Simplify inside:

1+3n1+n+n2.\frac{1+3n}{1+n+n^2}.

For large nn,

3nn2∼3n→0.\frac{3n}{n^2}\sim \frac{3}{n}\to0.

Thus term approaches

cos⁡(0)=1.\cos(0)=1.

Since terms do not go to 0, series diverges.

Exercise 31
Question

Determine whether the alternating series

∑n=1∞(−1)n+1(n+67)2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(n+67)^{2}}

is convergent or divergent. If convergent, show if absolutely or conditionally convergent.

Show solution ↓
Solution

Consider absolute series:

∑1(n+67)2.\sum \frac1{(n+67)^2}.

Since p=2>1, it converges.

Thus the alternating series is absolutely convergent.

Exercise 32
Question

Determine whether the alternating series

∑n=1∞(−1)n1(n−1)!(167)n\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{(n-1)!(\frac{1}{67})^{n}}

is convergent or divergent.

Show solution ↓
Solution

Rewrite:

1(n−1)!⋅67n.\frac{1}{(n-1)!}\cdot 67^n.

Using Ratio Test:

an+1an=67n.\frac{a_{n+1}}{a_n} = \frac{67}{n}.

Limit →0\to0.

Thus absolutely convergent.

Exercise 33
Question

Determine whether the alternating series

∑n=1∞(−1)n+1(n+1)!67n+1\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(n+1)!}{67^{n+1}}

is convergent or divergent.

Show solution ↓
Solution

Use Ratio Test:

an+1an=n+267.\frac{a_{n+1}}{a_n} = \frac{n+2}{67}.

Limit →∞\to\infty.

Thus diverges.

Exercise 34
Question

Determine whether the alternating series

∑n=1∞cos⁡(nπ)n+67\sum_{n=1}^{\infty}\frac{\cos(n\pi)}{n+67}

is convergent or divergent.

Show solution ↓
Solution
cos⁡(nπ)=(−1)n.\cos(n\pi)=(-1)^n.

Thus series is

∑(−1)nn+67.\sum \frac{(-1)^n}{n+67}.

Terms decrease to 0.

Thus converges by Alternating Series Test.

Absolute series diverges (harmonic type).

Hence conditionally convergent.

Exercise 35
Question

Find the radius and interval of convergence:

∑n=1∞xn5nn5.\sum_{n=1}^{\infty}\frac{x^n}{5^n n^5}.
Show solution ↓
Solution

Rewrite:

∑1n5(x5)n.\sum \frac1{n^5}\left(\frac{x}{5}\right)^n.

Ratio Test gives

\left|\frac{x}{5}\right|<1.

Thus radius R=5R=5.

Interval:

-5<x<5.

At endpoints p=5>1, converges.

Interval:

[−5,5].[-5,5].
Exercise 36
Question

Find radius and interval of convergence:

∑n=1∞(−1)nx4n(2n)!.\sum_{n=1}^{\infty}(-1)^n\frac{x^{4n}}{(2n)!}.
Show solution ↓
Solution

Factorial dominates any power.

Ratio Test gives limit 0 for all xx.

Thus radius R=∞R=\infty.

Interval: (−∞,∞)(-\infty,\infty).

Exercise 37
Question

Find radius and interval:

∑n=1∞(2n+1)!(x−2)nn11.\sum_{n=1}^{\infty}\frac{(2n+1)!(x-2)^n}{n^{11}}.
Show solution ↓
Solution

Factorial dominates polynomial.

Ratio Test gives divergence for any x≠2x\ne2.

Thus R=0R=0.

Interval: {2}\{2\} only.

Exercise 38
Question

Find radius and interval:

∑n=1∞(−1)n+1(x−4)nn9n.\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-4)^n}{n9^n}.
Show solution ↓
Solution

Rewrite:

∑1n(x−49)n.\sum \frac1n\left(\frac{x-4}{9}\right)^n.

Thus

\left|\frac{x-4}{9}\right|<1.

So R=9R=9.

Interval:

-5<x<13.

Check endpoints:

At x=13x=13, alternating harmonic → converges.

At x=−5x=-5, harmonic → diverges.

Interval:

(−5,13].(-5,13].
Exercise 39
Question

Find the Taylor series of e2xe^{2x} about x=12x=\frac12.

Show solution ↓
Solution

Taylor formula about a=12a=\frac12:

f(x)=∑n=0∞f(n)(a)n!(x−a)n.f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n.

Here f(x)=e2xf(x)=e^{2x}.

f(n)(x)=2ne2x.f^{(n)}(x)=2^n e^{2x}.

Thus

f(n)(12)=2ne1.f^{(n)}\left(\frac12\right) = 2^n e^{1}.

Therefore

e2x=∑n=0∞2nen!(x−12)n.e^{2x} = \sum_{n=0}^{\infty} \frac{2^n e}{n!}(x-\tfrac12)^n.
Exercise 40
Question

Use the Maclaurin series to show that

\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}, \quad -1<x\le1.
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Solution

We know

\frac1{1+x} = \sum_{n=0}^{\infty}(-1)^n x^n, \quad |x|<1.

Integrate both sides:

∫11+xdx=∫∑(−1)nxndx.\int \frac1{1+x}dx = \int \sum (-1)^n x^n dx.ln⁡(1+x)=∑(−1)nxn+1n+1.\ln(1+x) = \sum (-1)^n \frac{x^{n+1}}{n+1}.

Reindex:

ln⁡(1+x)=∑n=1∞(−1)n+1xnn.\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}.
Exercise 41
Question

Find the binomial series and interval of convergence of

f(x)=1(2+x)2.f(x)=\frac1{(2+x)^2}.
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Solution

Write:

1(2+x)2=141(1+x2)2.\frac1{(2+x)^2} = \frac1{4}\frac1{(1+\frac{x}{2})^2}.

Using binomial formula:

(1+u)−2=∑n=0∞(−1)n(n+1)un.(1+u)^{-2} = \sum_{n=0}^{\infty}(-1)^n (n+1)u^n.

Thus

1(2+x)2=14∑n=0∞(−1)n(n+1)(x2)n.\frac1{(2+x)^2} = \frac1{4} \sum_{n=0}^{\infty} (-1)^n(n+1)\left(\frac{x}{2}\right)^n.

Converges when

\left|\frac{x}{2}\right|<1.

Thus interval:

-2<x<2.
Exercise 42
Question

Find the period of f(x)=sin⁡(3x)f(x)=\sin(3x).

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Solution

For sin⁡(kx)\sin(kx), period is

2πk.\frac{2\pi}{k}.

Thus period is

2π3.\frac{2\pi}{3}.
Exercise 43
Question

Find the period of f(x)=cos⁡(5x−π4)f(x)=\cos(5x-\frac{\pi}{4}).

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Solution

Phase shift does not change period.

Period:

2π5.\frac{2\pi}{5}.
Exercise 44
Question

Find the period of f(x)=sin⁡(2x)f(x)=\sin(\sqrt2 x).

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Solution

Period:

2π2=2π.\frac{2\pi}{\sqrt2} = \sqrt2\pi.
Exercise 45
Question

Find the Fourier series of

f(x)={1,−πlt;xlt;02,0lt;xlt;πf(x)= \begin{cases} 1,&-\pi<x<0\\ 2,&0<x<\pi \end{cases}

and f(x+2π)=f(x)f(x+2\pi)=f(x)..

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Solution
a0=1π∫−ππf(x)dx=1π(π+2π)=3.a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)dx = \frac1\pi(\pi+2\pi) = 3.

Thus constant term a02=32\frac{a_0}{2}=\frac32.

an=1π∫−ππf(x)cos⁡(nx)dx=1π[1−2(−1)nnsin⁡(nπ)]=0.a_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos(nx)dx = \frac{1}{\pi}\left[\frac{1-2(-1)^n}{n}\sin(n\pi)\right] =0.bn=1π∫−ππf(x)sin⁡(nx)dx=1π1−(−1)nn.b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin(nx)dx = \frac{1}{\pi}\frac{1-(-1)^n}{n}.

Thus

f(x)=32+∑n=1∞1−(−1)nnπsin⁡(nx).f(x) = \frac32 + \sum_{n=1}^{\infty} \frac{1-(-1)^n}{n\pi}\sin(nx).
Exercise 46
Question

Check whether f(x)=∣x∣f(x)=|x| is odd, even, or neither.

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Solution
f(−x)=∣−x∣=∣x∣.f(-x)=|-x|=|x|.

Thus even.

Exercise 47
Question

Check whether f(x)=x3+4xf(x)=x^{3}+4x is odd, even, or neither.

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Solution
f(−x)=−x3−4x=−f(x).f(-x)=-x^3-4x=-f(x).

Thus odd.

Exercise 48
Question

Check whether f(x)=cos⁡xf(x)=\cos x is odd, even, or neither.

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Solution
cos⁡(−x)=cos⁡x.\cos(-x)=\cos x.

Thus even.

Exercise 49
Question

Check whether f(x)=x2+xf(x)=x^2+x is odd, even, or neither.

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Solution
f(−x)=x2−x.f(-x)=x^2-x.

Not equal to f(x)f(x) or −f(x)-f(x).

Thus neither.

Exercise 50
Question

Find Fourier series of f(x)=1+∣x∣f(x)=1+|x|, −π≤x≤π-\pi\le x\le\pi.

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Solution

Function is even.

Thus only cosine terms.

a0=2π∫0π(1+x)dx=2+π.a_0=\frac2\pi\int_0^\pi (1+x)dx = 2+\pi.an=2π∫0π(1+x)cos⁡(nx)dx=4(−1)nn2.a_n=\frac2\pi\int_0^\pi (1+x)\cos(nx)dx = \frac{4(-1)^n}{n^2}.

Thus

f(x)=1+π2+∑n=1∞4(−1)nn2cos⁡(nx).f(x) = 1+\frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2}\cos(nx).
Exercise 51
Question

Find the Fourier series of

f(x)={1,−πlt;xlt;00,0lt;xlt;πf(x)= \begin{cases} 1,&-\pi<x<0\\ 0,&0<x<\pi \end{cases}

and determine its value at x=0x=0.

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Solution
a0=1π∫−ππf(x)dx=1.a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)dx = 1.

Thus constant term 12\frac12.

bn=1π∫−π0sin⁡(nx)dx=1−(−1)nnπ.b_n=\frac1\pi\int_{-\pi}^0 \sin(nx)dx = \frac{1-(-1)^n}{n\pi}.

Thus

f(x)=12+∑n=1∞1−(−1)nnπsin⁡(nx).f(x) = \frac12 + \sum_{n=1}^{\infty} \frac{1-(-1)^n}{n\pi}\sin(nx).

At x=0x=0, Fourier series equals average of left and right limits:

1+02=12.\frac{1+0}{2}=\frac12.