Exercises to see if certain series converge.

Question

1. Does the series $\sum \frac{2^k+1}{3^k}$ converge?
2. Does the series $\sum (-1)^k\frac{1}{k^2}$ converge?
3. Does the series $\sum \frac{2^k+2^{2k}}{4^k}$ converge?
4. Does the series $\sum \frac{k!}{(2k)!}$ converge?

For the first question, do we use the comparison test comparing it to $\left(\frac{2}{3}\right)^k$???

For the second question, this series does converge. To prove this observe that this converges absolutely because \begin{equation*} \sum_{k=1}^{\infty} \left|(-1)^k\frac{1}{k^2}\right| = \sum_{k=1}^{\infty} \frac{1}{k^2} \end{equation*} converges. To see that this does converge, we can use the integral test.

The function $f:[1,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x^2}$ is continuous, non-negative and decreasing by examining the derivative $f'(x) = -\frac{2}{x^3} < 0$ for all $x\geq 1$. So we have \begin{equation*} \begin{split} \int_{1}^{\infty} \frac{1}{x^2} \; dx &= \left[-\frac{1}{x}\right]_{1}^{\infty} \\ &= \lim_{a\to\infty} \left(-\frac{1}{a}+1\right) \\ &= 1 \end{split} \end{equation*} This integral converges so by the integral test $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges as well and hence, $\sum_{k=1}^{\infty} (-1)^k\frac{1}{k^2}$ converges.

For the third question, this series does NOT converge. To show that this series diverges, we just need to show that \begin{equation*} \lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} \neq 0 \end{equation*} or doesn't exist. Well, for $k\geq 0$, \begin{equation*} \frac{2^k+2^{2k}}{4^k} = \frac{2^k+4^k}{4^k} = \left(\frac{1}{2}\right)^k+1. \end{equation*} Thus, \begin{equation*} \begin{split} \lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} &= \lim_{k\to\infty} \left(\frac{1}{2}\right)^k+1 \\ &= 0+1 \\ &= 1 \neq 0. \end{split} \end{equation*}

For the fourth question, if we let $a_k = \frac{k!}{(2k)!}$ then using the ratio test we get \begin{equation*} \begin{split} \lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty} \left|\frac{\frac{(k+1)!}{(2k+1)!}}{\frac{k!}{(2k)!}}\right| \\ &= \lim_{k\to\infty} \left|\frac{(k+1)!}{(2k+1)!}\times \frac{(2k)!}{k!}\right| \\ &= \lim_{k\to\infty} \left|\frac{(k+1)\times k\times (k-1)\times\ldots\times (2k)\times (2k-1)\times (2k-2)\times\ldots}{(2k+1)\times 2k\times (2k-1)\times\ldots\times k\times (k-1)\times (k-2)\times\ldots}\right| \\ &= \lim_{k\to\infty} \frac{k+1}{2k+1} \\ &= \lim_{k\to\infty} \frac{1+\frac{1}{k}}{2+\frac{1}{k}} \\ &= \frac{1}{2} < 1. \end{split} \end{equation*} Hence, $\sum_{k=0}^{\infty} \frac{k!}{(2k)!}$ converges.

Answer 1

As regards the last one, this is an alternative way: for $k\geq 1$, $$\frac{k!}{(2k)!}=\frac{1}{(2k)\cdot(2k-1)\cdots (k+1)}\leq \frac{1}{(2)\cdot(2)\cdots (2)}=\frac{1}{2^k}$$ and therefore the non-negative series $\sum \frac{k!}{(2k)!}$ converges by the comparison test.

Your answers for 2) and 3) are correct. For the first one, $$\frac{2^k+1}{3^k}\leq \frac{2^k+2^k}{3^k}=2\left(\frac{2}{3}\right)^k$$ and use the comparison test, or you may note that $$\sum \frac{2^k+1}{3^k}=\sum \left(\frac{2}{3}\right)^k+\sum \frac{1}{3^k}$$ which are both convergent.

Answer 2

For 1) use the fact that $\sum (\frac 2 3)^{k}$ and $\sum \frac1 {3^{k}}$ both converge.

Your answer for 2) and 3) are OK.

In 4) $a_{k+1}$ is not $\frac {(k+1)!} {(2k+1)!}$. It is $\frac {(k+1)!} {(2k+2)!}$. [You have to replace $k$ by $k+1$ in the formula for $a_k$]. But your approach is good and you will be able to complete it after making this correction.