Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

I need to be able to prove that this series converges. I know I need to use the ratio test but I do not know how to go about doing it.

Any help is much appreciated! thank you

share|cite|improve this question
What about the root test? Do you know this one? Can you use it? – 1015 Apr 28 '13 at 13:48

It’s possible to use the ratio test, though it probably isn’t the easiest approach:

$$\begin{align*} \frac{\left(\frac{k+1}{k+2}\right)^{(k+1)^2}}{\left(\frac{k}{k+1}\right)^{k^2}}&=\frac{(k+1)^{2k^2}}{k^{k^2}(k+2)^{k^2}}\cdot\left(\frac{k+1}{k+2}\right)^{2k+1}\\\\ &=\left(\frac{k^2+2k+1}{k^2+2k}\right)^{k^2}\cdot\left(\frac{k+1}{k+2}\right)^{2k+1}\\\\ &=\left(1+\frac1{k^2+2k}\right)^{k^2}\cdot\left(1-\frac1{k+2}\right)^{2k+1}\\\\ &=\left(\left(1+\frac1{k^2+2k}\right)^{k^2+2k}\right)^{\frac{k}{k+2}}\cdot\left(\left(1-\frac1{k+2}\right)^{k+2}\right)^{\frac{2k+1}{k+2}}\;. \end{align*}$$

For large $k$ the quantities

$$\left(1+\frac1{k^2+2k}\right)^{k^2+2k}\quad\text{ and }\quad\left(1-\frac1{k+2}\right)^{k+2}$$

have approximately what numerical values?

share|cite|improve this answer

In fact \begin{eqnarray*} L&=&\lim_{k\to\infty}\frac{a_{k+1}}{a_k}=\lim_{k\to\infty}\left(\frac{k+1}{k+2}\right)^{(k+1)^2}\left(\frac{k+1}{k}\right)^{k^2}\\ &=&\lim_{k\to\infty}\left(\frac{k+1}{k+2}\right)^{k^2}\left(\frac{k+1}{k}\right)^{k^2}\left(\frac{k+1}{k+2}\right)^{2k+1}\\ &=&\lim_{k\to\infty}\left(\frac{(k+1)^2}{k(k+2)}\right)^{k^2}\left(1-\frac{1}{k+2}\right)^{2k+1}\\ &=&\lim_{k\to\infty}\left(1+\frac{1}{k(k+2)}\right)^{k^2}\left(1-\frac{1}{k+2}\right)^{2k+1}\\ &=&\lim_{k\to\infty}\left(1+\frac{1}{k(k+2)}\right)^{k(k+2)\frac{k^2}{k(k+2)}}\left(1-\frac{1}{k+2}\right)^{(k+2)\frac{2k+1}{k+2}}\\ &=&e\times\frac{1}{e^2}=\frac{1}{e}<1. \end{eqnarray*} So the series converges. You also can use the root test to get $L=\frac{1}{e}$ which is much easier to calculate.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.