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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

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  • $\begingroup$ What about the root test? Do you know this one? Can you use it? $\endgroup$
    – Julien
    Apr 28, 2013 at 13:48

2 Answers 2

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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?

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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.

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