Exercise 1

Use mathematical induction to prove:

1^{2}+2^{2}+3^{2}+\dots+n^{2} = \frac{n(n+1)(2n+1)}{6}

for any positive integer $n$ .

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Solution

Step 1: Base Case

For $n=1$ :

1^{2}=1

\frac{1(1+1)(2\cdot1+1)}{6} = \frac{1\cdot2\cdot3}{6} = 1

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

Step 2: Inductive Hypothesis

Assume the formula holds for $n=k$ :

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

Step 3: Inductive Step

For $n=k+1$ :

1^{2}+2^{2}+\dots+k^{2}+(k+1)^{2}.

Using the hypothesis:

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

Factor $(k+1)$ :

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

Combine into one fraction:

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

Simplify numerator:

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

Thus,

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

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

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

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

Conclusion

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

Exercise 2

Use mathematical induction to prove:

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

for any positive integer $n$ .

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Solution

Base Case: For $n=1$ ,

3^{1}-1=2,

which is divisible by 2.

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

Inductive Step:

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

Since $3^{k}-1=2m$ ,

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

Thus divisible by 2.

Therefore, true for all $n$ .

Exercise 3

Use mathematical induction to prove:

a+ar+ar^{2}+\dots+ar^{n-1} = \frac{a(1-r^{n})}{1-r}, \quad r\ne1.

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Solution

Base Case: For $n=1$ ,

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

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

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

Inductive Step: Add $ar^{k}$ :

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

Write over common denominator:

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

Simplify numerator:

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

Thus,

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

True for all $n$ .

Exercise 4

Use mathematical induction to prove:

2^{n}\ge n+1

for any positive integer $n$ .

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Solution

Base Case: For $n=1$ ,

2^{1}=2 \ge 2.

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

Inductive Step:

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

Since $2k+2 \ge k+2$ for $k\ge1$ ,

2^{k+1}\ge k+2.

Thus true for all $n$ .

Exercise 5

Write the first four terms of the sequence:

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

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Solution

For $n=1$ :

(-1)\frac{2}{1}= -2.

$n=2$ :

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

$n=3$ :

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

$n=4$ :

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

First four terms:

-2,\;2,\;-\frac{4}{3},\;\frac{2}{3}.

Exercise 6

Write the first four terms of the sequence:

\left\{\sin\left(\frac{n\pi}{2}\right)-\cos(n\pi)\right\}_{n=0}^{\infty}.

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Solution

$n=0$ :

0-1=-1.

$n=1$ :

1-(-1)=2.

$n=2$ :

0-1=-1.

$n=3$ :

-1-(-1)=0.

First four terms:

-1,\;2,\;-1,\;0.

Exercise 7

Determine whether the sequence

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

is convergent or divergent. Find its limit if convergent.

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Solution

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

As $n\to\infty$ , $\frac{1}{n}\to0$ .

Thus,

\lim_{n\to\infty}\frac{n-1}{n}=1.

The sequence converges to 1.

Exercise 8

Determine whether the sequence

\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.

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Solution

Simplify:

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

Thus,

\frac{n}{\sqrt{n^2+1}}\cdot\frac{1}{n} = \frac{1}{\sqrt{1+\frac{1}{n^2}}}.

As $n\to\infty$ ,

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

The sequence converges to 1.

Exercise 9

Determine whether the sequence

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

is convergent or divergent. Find its limit if convergent.

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Solution

Rewrite:

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

Thus,

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

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

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

We know

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

Therefore,

\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 $\frac{1}{e}$ .

Exercise 10

Determine whether the sequence

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

is convergent or divergent.

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Solution

As $n\to\infty$ , $2^n$ grows much faster than $n^2$ .

Thus,

\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

Determine whether the sequence

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

is convergent or divergent. Find its limit if convergent.

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Solution

We know

-1\le \cos n \le 1.

Thus

-\frac{1}{\sqrt n}\le \frac{\cos n}{\sqrt n} \le \frac{1}{\sqrt n}.

Since $\frac{1}{\sqrt n}\to0$ ,

\frac{\cos n}{\sqrt n}\to0.

Therefore,

4+\frac{\cos n}{\sqrt n}\to4.

The sequence converges to 4.

Exercise 12

Determine whether the sequence

\{n-n^{2}\}_{n=1}^{\infty}

is monotonic.

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Solution

a_n=n-n^2=-n^2+n.

Compute difference:

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

Simplify:

= -2n.

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

Thus it is monotonic decreasing.

Exercise 13

Determine whether the sequence

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

is monotonic.

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Solution

Consider ratio:

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

For $n\ge1$ ,

\frac{2}{n+2}&lt;1.

Thus $a_{n+1}<a_n$ .

The sequence is decreasing.

Exercise 14

Determine whether the sequence

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

is monotonic.

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Solution

a_n=\frac{n}{n+1}.

Compute difference:

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)}&gt;0.

Thus increasing.

The sequence is monotonic increasing.

Exercise 15

Determine whether the series

\sum_{n=1}^{\infty}\frac{(-3)^{n}}{2^{2n}}

is convergent or divergent. Find its sum if convergent.

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Solution

2^{2n}=4^n.

Thus series becomes

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

This is geometric with ratio $r=-\frac34$ .

Since $|r|<1$ , it converges.

Sum:

\frac{r}{1-r} = \frac{-\frac34}{1+\frac34} = -\frac{3}{7}.

Exercise 16

Determine whether the series

\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:

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

This telescopes.

Sum becomes

32\left(1-\lim_{n\to\infty}\frac1{n+1}\right) =32.

Thus convergent with sum $32$ .

Exercise 17

Determine whether the series

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

is convergent or divergent. Find its sum if convergent.

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Solution

First simplify the term.

3^{1-n}=\frac{3}{3^n}.

Thus,

\frac{23(2^{n-1})}{3^{1-n}} = 23(2^{n-1})\frac{3^n}{3} = \frac{23}{3}2^{n-1}3^n.

= \frac{23}{3}\cdot \frac{2^n}{2}\cdot 3^n = \frac{23}{6}(6^n).

Thus the series becomes

\sum \frac{23}{6}6^n.

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

Therefore the series diverges.

Exercise 18

Determine whether the series

\sum_{n=3}^{\infty}\frac{3}{n^{2}-2n}

is convergent or divergent. Find its sum if convergent.

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Solution

Factor denominator:

n^2-2n=n(n-2).

Use partial fractions:

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

Solving gives:

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

Thus series telescopes:

\frac{3}{2}\left(\frac1{1}+\frac1{2}-\lim_{n\to\infty}\frac1n-\frac1{n-1}\right).

Remaining terms:

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

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

Exercise 19

Determine whether the series

\sum_{n=1}^{\infty}n^{3n}\sin(2n\pi)

is convergent or divergent.

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Solution

We know

\sin(2n\pi)=0

for all integers $n$ .

Thus every term is zero.

So the series equals 0.

Therefore it converges (sum = 0).

Exercise 20

Find the value of $a$ which makes the series

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

convergent.

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Solution

This is a geometric series:

\sum (-1)^n(1+a)^n.

Common ratio:

r=-(1+a).

Geometric series converges if

|r|&lt;1.

Thus

|1+a|&lt;1.

Solve:

-1&lt;1+a&lt;1.

-2&lt;a&lt;0.

Exercise 21

Use the integral test to show that

\sum_{n=1}^{\infty}\frac{1}{n^{p}}

is convergent when $p>1$ .

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Solution

Consider

\int_1^\infty \frac1{x^p}dx.

= \int_1^\infty x^{-p}dx = \left[\frac{x^{1-p}}{1-p}\right]_1^\infty.

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

Thus $x^{1-p}\to0$ as $x\to\infty$ .

So integral is finite.

Therefore the series converges for $p>1$ .

Exercise 22

Determine whether the series

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

is convergent or divergent.

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Solution

\frac{1}{n+2}\sim\frac1n.

Since harmonic series diverges,

\sum \frac1{n+2}

also diverges by comparison.

Exercise 23

Determine whether the series

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

is convergent or divergent.

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Solution

Since

2^n+n^2&gt;2^n,

\frac1{2^n+n^2}&lt;\frac1{2^n}.

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

Thus by comparison, given series converges.

Exercise 24

Determine whether the series

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

is convergent or divergent.

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Solution

Use Ratio Test.

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

As $n\to\infty$ , ratio $\to\infty$ .

Thus series diverges.

Exercise 25

Determine whether the series

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

is convergent or divergent.

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Solution

Factorial growth in numerator is much faster than denominator.

Terms grow rapidly and do not approach zero.

Therefore the series diverges.

Exercise 26

Determine whether the series

\sum_{n=1}^{\infty}\frac{\sqrt{\ln(n^{3})}}{3n}

is convergent or divergent.

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Solution

\ln(n^3)=3\ln n.

Thus term behaves like

\frac{\sqrt{\ln n}}{n}.

Since

\frac{\sqrt{\ln n}}{n} &gt; \frac1n \text{ eventually},

and harmonic diverges,

the series diverges.

Exercise 27

Determine whether the series

\sum_{n=2}^{\infty}\frac{n}{(n^{3}+3)(3n-3)}

is convergent or divergent.

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Solution

For large $n$ ,

(n^{3}+3)(3n-3)\sim 3n^4.

Thus

\frac{n}{(n^{3}+3)(3n-3)} \sim \frac{n}{3n^4} = \frac{1}{3n^3}.

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

Exercise 28

Determine whether the series

\sum_{n=3}^{\infty}\frac{1}{3^{-n}(n-3)!}

is convergent or divergent.

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Solution

\frac{1}{3^{-n}(n-3)!} = \frac{3^n}{(n-3)!}.

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

Using Ratio Test:

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

As $n\to\infty$ , ratio $\to0$ .

Thus series converges absolutely.

Exercise 29

Determine whether the series

\sum_{n=1}^{\infty}\frac{2\cdot3^{3n}+3^{n}}{3-3^{3n+1}}

is convergent or divergent.

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Solution

Dominant term is $3^{3n}$ .

Expression behaves like

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

Terms do not approach 0.

Thus the series diverges.

Exercise 30

Determine whether the series

\sum_{n=1}^{\infty}\cos\left(\frac{1+n+2n}{1+n+n^{2}}\right)

is convergent or divergent.

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Solution

Simplify inside:

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

For large $n$ ,

\frac{3n}{n^2}\sim \frac{3}{n}\to0.

Thus term approaches

\cos(0)=1.

Since terms do not go to 0, series diverges.

Exercise 31

Determine whether the alternating series

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

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

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Solution

Consider absolute series:

\sum \frac1{(n+67)^2}.

Since $p=2>1$ , it converges.

Thus the alternating series is absolutely convergent.

Exercise 32

Determine whether the alternating series

\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{(n-1)!(\frac{1}{67})^{n}}

is convergent or divergent.

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Solution

Rewrite:

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

Using Ratio Test:

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

Limit $\to0$ .

Thus absolutely convergent.

Exercise 33

Determine whether the alternating series

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

is convergent or divergent.

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Solution

Use Ratio Test:

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

Limit $\to\infty$ .

Thus diverges.

Exercise 34

Determine whether the alternating series

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

is convergent or divergent.

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Solution

\cos(n\pi)=(-1)^n.

Thus series is

\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

Find the radius and interval of convergence:

\sum_{n=1}^{\infty}\frac{x^n}{5^n n^5}.

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Solution

Rewrite:

\sum \frac1{n^5}\left(\frac{x}{5}\right)^n.

Ratio Test gives

\left|\frac{x}{5}\right|&lt;1.

Thus radius $R=5$ .

Interval:

-5&lt;x&lt;5.

At endpoints $p=5>1$ , converges.

Interval:

[-5,5].

Exercise 36

Find radius and interval of convergence:

\sum_{n=1}^{\infty}(-1)^n\frac{x^{4n}}{(2n)!}.

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Solution

Factorial dominates any power.

Ratio Test gives limit 0 for all $x$ .

Thus radius $R=\infty$ .

Interval: $(-\infty,\infty)$ .

Exercise 37

Find radius and interval:

\sum_{n=1}^{\infty}\frac{(2n+1)!(x-2)^n}{n^{11}}.

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Solution

Factorial dominates polynomial.

Ratio Test gives divergence for any $x\ne2$ .

Thus $R=0$ .

Interval: $\{2\}$ only.

Exercise 38

Find radius and interval:

\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-4)^n}{n9^n}.

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Solution

Rewrite:

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

Thus

\left|\frac{x-4}{9}\right|&lt;1.

So $R=9$ .

Interval:

-5&lt;x&lt;13.

Check endpoints:

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

At $x=-5$ , harmonic → diverges.

Interval:

(-5,13].

Exercise 39

Find the Taylor series of $e^{2x}$ about $x=\frac12$ .

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Solution

Taylor formula about $a=\frac12$ :

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

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

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

Thus

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

Therefore

e^{2x} = \sum_{n=0}^{\infty} \frac{2^n e}{n!}(x-\tfrac12)^n.

Exercise 40

Use the Maclaurin series to show that

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

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Solution

We know

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

Integrate both sides:

\int \frac1{1+x}dx = \int \sum (-1)^n x^n dx.

\ln(1+x) = \sum (-1)^n \frac{x^{n+1}}{n+1}.

Reindex:

\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}.

Exercise 41

Find the binomial series and interval of convergence of

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

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Solution

Write:

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

Using binomial formula:

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

Thus

\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|&lt;1.

Thus interval:

-2&lt;x&lt;2.

Exercise 42

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

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Solution

For $\sin(kx)$ , period is

\frac{2\pi}{k}.

Thus period is

\frac{2\pi}{3}.

Exercise 43

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

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Solution

Phase shift does not change period.

Period:

\frac{2\pi}{5}.

Exercise 44

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

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Solution

Period:

\frac{2\pi}{\sqrt2} = \sqrt2\pi.

Exercise 45

Find the Fourier series of

f(x)= \begin{cases} 1,&-\pi&lt;x&lt;0\\ 2,&0&lt;x&lt;\pi \end{cases}

and $f(x+2\pi)=f(x)$ ..

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Solution

a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)dx = \frac1\pi(\pi+2\pi) = 3.

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

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.

b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin(nx)dx = \frac{1}{\pi}\frac{1-(-1)^n}{n}.

Thus

f(x) = \frac32 + \sum_{n=1}^{\infty} \frac{1-(-1)^n}{n\pi}\sin(nx).

Exercise 46

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

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Solution

f(-x)=|-x|=|x|.

Thus even.

Exercise 47

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

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Solution

f(-x)=-x^3-4x=-f(x).

Thus odd.

Exercise 48

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

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Solution

\cos(-x)=\cos x.

Thus even.

Exercise 49

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

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Solution

f(-x)=x^2-x.

Not equal to $f(x)$ or $-f(x)$ .

Thus neither.

Exercise 50

Find Fourier series of $f(x)=1+|x|$ , $-\pi\le x\le\pi$ .

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Solution

Function is even.

Thus only cosine terms.

a_0=\frac2\pi\int_0^\pi (1+x)dx = 2+\pi.

a_n=\frac2\pi\int_0^\pi (1+x)\cos(nx)dx = \frac{4(-1)^n}{n^2}.

Thus

f(x) = 1+\frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2}\cos(nx).

Exercise 51

Find the Fourier series of

f(x)= \begin{cases} 1,&-\pi&lt;x&lt;0\\ 0,&0&lt;x&lt;\pi \end{cases}

and determine its value at $x=0$ .

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Solution

a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)dx = 1.

Thus constant term $\frac12$ .

b_n=\frac1\pi\int_{-\pi}^0 \sin(nx)dx = \frac{1-(-1)^n}{n\pi}.

Thus

f(x) = \frac12 + \sum_{n=1}^{\infty} \frac{1-(-1)^n}{n\pi}\sin(nx).

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

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

Exercise Set 1

Theory

Exercises