Series Practice

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1.1Necessary Condition for Convergence

Exam I | Problem 1.1 |

Necessary condition for convergence

If the series

\sum_{n=1}^{\infty} a_n

converges, what must

\lim_{n\to\infty} a_n

be?

Solution

If a series converges, its terms must go to $0$.

\lim_{n\to\infty} a_n = 0.

This condition is necessary, but not sufficient.

1.1Find a Radius of Convergence

Exam I | Problem 1.1 |

Find the radius of convergence of

\sum_{n=0}^{\infty}\frac{(x-2)^n}{3^n(n+1)}.

Solution

The ratio test gives $|x-2|/3$, so convergence requires $|x-2|<3$. The radius is $R=3$; test the endpoints separately.

1.2Sum a Geometric Series

Exam I | Problem 1.2 |

Geometric series

Find the sum of

\sum_{n=0}^{\infty} 6\left(\frac{1}{4}\right)^n.

Solution

This is a geometric series with first term $a=6$ and ratio $r=\frac14$.

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

\sum_{n=0}^{\infty} 6\left(\frac{1}{4}\right)^n = \frac{6}{1-\frac14} = \frac{6}{\frac34} = 8.

1.2Test Power-Series Endpoints

Exam I | Problem 1.2 |

For the previous series, test the endpoints.

Solution

At $x=5$, the terms become $1/(n+1)$ and diverge. At $x=-1$, they become $(-1)^n/(n+1)$ and converge by the alternating series test. The interval is $[-1,5)$.

1.3Compute a Finite Geometric Sum

Exam I | Problem 1.3 |

Finite geometric sum

Compute

\sum_{n=0}^{3} 2\left(\frac12\right)^n.

Solution

Use the finite geometric sum formula:

\sum_{n=0}^{N-1} ar^n = a\frac{1-r^N}{1-r}.

Here $a=2$, $r=\frac12$, and $N=4$:

\sum_{n=0}^{3} 2\left(\frac12\right)^n = 2\cdot \frac{1-(\frac12)^4}{1-\frac12} = 2\cdot \frac{1-\frac1{16}}{\frac12} = \frac{15}{4}.

1.3Write a Taylor Polynomial

Exam I | Problem 1.3 |

Find the degree-two Taylor polynomial for $f(x)=\ln x$ centered at $a=1$.

Solution

Since $f(1)=0$, $f'(1)=1$, and $f''(1)=-1$,

P_2(x)=(x-1)-\frac12(x-1)^2.

1.4Evaluate a Telescoping Sum

Exam I | Problem 1.4 |

Telescoping series

Evaluate

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

Solution

Write out the partial sums:

s_N = \left(1-\frac13\right) + \left(\frac12-\frac14\right) + \cdots + \left(\frac1N-\frac1{N+2}\right).

Most terms cancel, leaving

s_N = 1+\frac12-\frac{1}{N+1}-\frac{1}{N+2}.

Now let $N\to\infty$:

\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right) = 1+\frac12 = \frac32.

1.4Use a Taylor Approximation

Exam I | Problem 1.4 |

Use $1+x+x^2/2$ to approximate $e^{0.2}$ and state the next-term estimate.

Solution

The approximation is $1+0.2+0.2^2/2=1.22$. The next term is $0.2^3/6\approx0.00133$, which is a rough local error scale, not an unconditional bound.

1.5Decide a p-Series

Exam I | Problem 1.5 |

p-series

Does

\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}

converge or diverge?

Solution

This is a $p$-series with

p=\frac32.

Since $p>1$, the series converges.

1.5Derive a Logarithm Series

Exam I | Problem 1.5 |

Starting with $1/(1-x)=\sum_{n=0}^{\infty}x^n$, derive a series for $\ln(1+x)$.

Solution

Replace $x$ by $-x$ and integrate term by term:

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

1.6Classify an Alternating p-Series

Exam I | Problem 1.6 |

Classify the series

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

as absolutely convergent, conditionally convergent, or divergent.

Solution

First check absolute convergence:

\sum_{n=1}^{\infty}\left|(-1)^{n+1}\frac{1}{\sqrt{n}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{1/2}}.

This is a $p$-series with $p=\frac12$, so it diverges.

Now apply the alternating series test to

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

The terms decrease to $0$, so the series converges.

Therefore, the series is conditionally convergent.

1.6Match Power-Series Coefficients

Exam I | Problem 1.6 |

If $\sum_{n=0}^{\infty}a_nx^n=e^x$, find $a_0$, $a_1$, and $a_2$.

Solution

Comparing with $e^x=1+x+x^2/2!+\cdots$ gives $a_0=1$, $a_1=1$, and $a_2=1/2$.

1.7Spot Divergence from the Terms

Exam I | Problem 1.7 |

Divergence test

Does

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

converge or diverge?

Solution

Use the divergence test by checking the term limit:

\lim_{n\to\infty}\frac{2n+1}{n+2} = 2.

Since the terms do not go to $0$, the series diverges.

1.7Identify Conditional Convergence

Exam I | Problem 1.7 |

Classify $\sum_{n=1}^{\infty}(-1)^{n+1}/n$.

Solution

It converges by the alternating series test, but its absolute series is harmonic and diverges. The convergence is conditional.

1.8Compare to a Known Convergent Series

Exam I | Problem 1.8 |

Comparison test

Does

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

converge or diverge?

Solution

For $n\ge 1$,

0<\frac{1}{n^2+1}\le \frac{1}{n^2}.

Since

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

converges, the comparison test shows that

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

also converges.

1.8Interpret a Fourier Preview

Exam I | Problem 1.8 |

What does the coefficient multiplying $\cos(nx)$ represent qualitatively?

Solution

It measures the contribution of the cosine mode with frequency $n$ to the periodic signal. It is an amplitude determined by the function and interval, not itself a frequency or probability.

1.9Geometric Series with a Negative Ratio

Exam I | Problem 1.9 |

Geometric series

Find the sum of

\sum_{n=0}^{\infty} 5\left(-\frac34\right)^n.

Solution

This is geometric with first term $a=5$ and ratio $r=-\frac34$.

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

\sum_{n=0}^{\infty} 5\left(-\frac34\right)^n = \frac{5}{1-(-\frac34)} = \frac{5}{\frac74} = \frac{20}{7}.

1.10Check Absolute Convergence

Exam I | Problem 1.10 |

Absolute convergence check

Determine whether

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

is absolutely convergent, conditionally convergent, or divergent.

Solution

Check the absolute series:

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

This is a $p$-series with $p=2$, so it converges.

Therefore, the original series converges absolutely.

Difficulty

Integrated

2.1Use Direct Comparison

Exam II | Problem 2.1 |

Comparison test

Determine whether

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

converges or diverges.

Solution

For $n\ge 1$,

0<\frac{1}{n^2+3n}\le \frac{1}{n^2}.

Since $\sum \frac{1}{n^2}$ converges, the comparison test gives convergence of the given series.

2.2Use Limit Comparison

Exam II | Problem 2.2 |

Limit comparison test

Determine whether

\sum_{n=1}^{\infty}\frac{4n+1}{n^2+n}

converges or diverges.

Solution

Compare with

b_n=\frac1n.

Then

\lim_{n\to\infty}\frac{\frac{4n+1}{n^2+n}}{\frac1n} = \lim_{n\to\infty}\frac{4n^2+n}{n^2+n} = 4.

Because the limit is a positive finite number, the two series behave the same way.

Since

\sum_{n=1}^{\infty}\frac1n

diverges, the given series also diverges.

2.3Apply the Integral Test

Exam II | Problem 2.3 |

Advanced convergence test: integral test

Determine whether

\sum_{n=1}^{\infty}\frac{1}{n^2+4}

converges or diverges using the integral test.

Solution

Let

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

For $x\ge 1$, this function is positive, continuous, and decreasing, so the integral test applies.

Compute the improper integral:

\int_1^{\infty}\frac{1}{x^2+4}\,dx = \frac12\int_1^{\infty}\frac{1}{(x/2)^2+1}\,dx = \frac12\left[\arctan\!\left(\frac{x}{2}\right)\right]_1^{\infty}.

This is finite, so the series converges.

2.4Apply the Ratio Test

Exam II | Problem 2.4 |

Ratio test

Determine whether

\sum_{n=0}^{\infty}\frac{n!}{3^n}

converges or diverges.

Solution

Let

a_n=\frac{n!}{3^n}.

Then

\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{3^{n+1}}\cdot \frac{3^n}{n!} = \frac{n+1}{3}.

As $n\to\infty$,

\frac{n+1}{3}\to\infty.

So the ratio test gives divergence.

2.5Apply the Root Test

Exam II | Problem 2.5 |

Root test

Determine whether

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

converges or diverges.

Solution

Use the root test:

L=\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{2n}{3n+1}\right)^n\right|} = \lim_{n\to\infty}\frac{2n}{3n+1} = \frac23.

Since $L<1$, the series converges absolutely.

2.6Find an Interval of Convergence

Exam II | Problem 2.6 |

Find the interval of convergence of

\sum_{n=1}^{\infty}\frac{(x-2)^n}{n4^n}.

Solution

Apply the ratio test to the general term

a_n=\frac{(x-2)^n}{n4^n}.

Then

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x-2}{4}\right|\cdot \frac{n}{n+1}.

The limit is

\left|\frac{x-2}{4}\right|.

So the series converges when

\left|\frac{x-2}{4}\right|<1,

which gives

$$ -2<x<6. $$

Check the endpoints:

At $x=6$,

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

diverges.

At $x=-2$,

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

converges conditionally by the alternating series test.

Therefore the interval of convergence is

$$ [-2,6). $$

2.7Differentiate a Power Series

Exam II | Problem 2.7 |

For $|x|<1$, find a power series for

\frac{x}{(1-x)^2}.

Solution

Start with the geometric series:

\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n,\quad |x|<1.

Differentiate term by term:

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

Multiply both sides by $x$:

\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty} n x^n,\quad |x|<1.

2.8Expand a Rational Function by Substitution

Exam II | Problem 2.8 |

For $|x|<1$, write

\frac{1}{1-x^2}

as a power series.

Solution

Start with the geometric series

\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n,\quad |u|<1.

Substitute $u=x^2$:

\frac{1}{1-x^2}=\sum_{n=0}^{\infty}x^{2n}.

This is valid when

$$ |x^2|<1, $$

so the interval of convergence is

$$ |x|<1. $$

Difficulty

Applied

3.1Reindex a Series

Final | Problem 3.1 |

Index shifting

Rewrite

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

in standard form, and decide whether it converges.

Solution

Let $k=n-1$. Then when $n=2$, $k=1$, and the series becomes

\sum_{k=1}^{\infty}\frac{1}{k^2}.

This is a $p$-series with $p=2$, so it converges.

3.2Choose the Right Comparison

Final | Problem 3.2 |

Limit comparison test

Determine whether

\sum_{n=1}^{\infty}\frac{n^2}{n^3+5}

converges or diverges.

Solution

Compare with

b_n=\frac1n.

Compute

\lim_{n\to\infty}\frac{\frac{n^2}{n^3+5}}{\frac1n} = \lim_{n\to\infty}\frac{n^3}{n^3+5} = 1.

So the series behaves like $\sum \frac1n$.

Because the harmonic series diverges, the given series diverges as well.

3.3Evaluate a Telescoping Series

Final | Problem 3.3 |

Telescoping series

Evaluate

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

Solution

Use partial fractions:

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.

So the partial sums telescope:

s_N=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1N-\frac{1}{N+1}\right).

Everything cancels except the first and last pieces:

s_N=1-\frac{1}{N+1}.

Taking the limit gives

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

3.4Build a Logarithm Series

Final | Problem 3.4 |

Use the geometric series to write the Maclaurin series for

-\ln(1-x).

State the interval where the series converges.

Solution

Start with

\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n,\quad |x|<1.

Integrate both sides term by term:

\int \frac{1}{1-x}\,dx = \int \sum_{n=0}^{\infty}x^n\,dx.

This gives

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

The first four nonzero terms are

x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}.

3.5Use a Power Series to Sum a Series

Final | Problem 3.5 |

Evaluate

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

Solution

Differentiate the geometric series:

\sum_{n=0}^{\infty}x^n=\frac{1}{1-x},\quad |x|<1

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

Now set $x=\frac12$:

\sum_{n=1}^{\infty}n\left(\frac12\right)^{n-1} = \frac{1}{(1-\frac12)^2} = 4.

Multiply both sides by $\frac12$:

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

Difficulty

Challenge

4.1Approximate with a Taylor Polynomial

Final | Problem 4.1 |

Use the Maclaurin series for $\sin x$ to approximate $\sin(0.2)$ with the first two nonzero terms.

Solution

The Maclaurin series for sine is

\sin x = x - \frac{x^3}{3!} + \cdots

Use the first two nonzero terms:

\sin(0.2)\approx 0.2-\frac{(0.2)^3}{6}.

Since

$$ 0.2^3=0.008, $$

we get

\sin(0.2)\approx 0.2-\frac{0.008}{6} = 0.198666\ldots = \frac{149}{750}.

The next omitted term is very small, so this is a good local approximation.

4.2Find a Binomial Coefficient

Final | Problem 4.2 |

Binomial-type expansion

In the expansion of

(1+x)^{1/2},

what is the coefficient of $x^3$?

Solution

Use the binomial-type expansion:

(1+x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n.

For $\alpha=\frac12$, the coefficient of $x^3$ is

\binom{1/2}{3} = \frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)}{3!} = \frac{1}{16}.

4.3Match Coefficients

Final | Problem 4.3 |

Matching coefficients

Suppose

f(x)=\sum_{n=0}^{\infty}a_n x^n

and

$$ (1-x)f(x)=1+x. $$

Find the coefficients $a_n$.

Solution

Expand the left-hand side:

(1-x)\sum_{n=0}^{\infty}a_n x^n = \sum_{n=0}^{\infty}a_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1}.

Reindex the second sum:

\sum_{n=0}^{\infty}a_n x^n - \sum_{n=1}^{\infty}a_{n-1}x^n.

(1-x)f(x)=a_0+\sum_{n=1}^{\infty}(a_n-a_{n-1})x^n.

Match coefficients with

$$ 1+x. $$

This gives

$$ a_0=1, $$

$$ a_1-a_0=1, $$

and for $n\ge 2$,

a_n-a_{n-1}=0.

$$ a_1=2 $$

and

a_n=2 \quad \text{for all } n\ge 1.

4.4Classify a Mixed-Sign Series

Final | Problem 4.4 |

Classify

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

as absolutely convergent, conditionally convergent, or divergent.

Solution

First check absolute convergence:

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

Since

\frac{n}{n^2+1}\sim \frac1n,

this absolute series diverges.

Now check the alternating series test. Let

b_n=\frac{n}{n^2+1}.

Then $b_n\to 0$, and $b_n$ is decreasing for $n\ge 1$ because the function

f(x)=\frac{x}{x^2+1}

has derivative

f'(x)=\frac{1-x^2}{(x^2+1)^2}\le 0

for $x\ge 1$.

So the alternating series converges, but not absolutely.

Therefore, it is conditionally convergent.