Series Practice

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

So

This condition is necessary, but not sufficient.

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

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

Use the finite geometric sum formula:

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

Write out the partial sums:

Most terms cancel, leaving

Now let $N\to\infty$:

This is a $p$-series with

Since $p>1$, the series converges.

First check absolute convergence:

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

Now apply the alternating series test to

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

Therefore, the series is conditionally convergent.

Use the divergence test by checking the term limit:

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

For $n\ge 1$,

Since

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

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

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

Check the absolute series:

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

Therefore, the original series converges absolutely.

For $n\ge 1$,

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

Compare with

Then

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

Since

diverges, the given series also diverges.

Let

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

Compute the improper integral:

This is finite, so the series converges.

Let

Then

As $n\to\infty$,

So the ratio test gives divergence.

Use the root test:

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

Apply the ratio test to the general term

Then

The limit is

So the series converges when

which gives

Check the endpoints:

At $x=6$,

diverges.

At $x=-2$,

converges conditionally by the alternating series test.

Therefore the interval of convergence is

Start with the geometric series:

Differentiate term by term:

Multiply both sides by $x$:

Start with the geometric series

Substitute $u=x^2$:

This is valid when

so the interval of convergence is

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

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

Compare with

Compute

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

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

Use partial fractions:

So the partial sums telescope:

Everything cancels except the first and last pieces:

Taking the limit gives

Start with

Integrate both sides term by term:

This gives

The first four nonzero terms are

Differentiate the geometric series:

so

Now set $x=\frac12$:

Multiply both sides by $\frac12$:

The Maclaurin series for sine is

Use the first two nonzero terms:

Since

we get

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

Use the binomial-type expansion:

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

Expand the left-hand side:

Reindex the second sum:

So

Match coefficients with

This gives

and for $n\ge 2$,

So

and

First check absolute convergence:

Since

this absolute series diverges.

Now check the alternating series test. Let

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

has derivative

for $x\ge 1$.

So the alternating series converges, but not absolutely.

Therefore, it is conditionally convergent.