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WolframAlpha tells me that the sum $$\displaystyle \sum_{n=1}^\infty\left(\frac{n}{n+1}\right)^{n^2}\tag1$$ converges by the ratio test, but I have absolutely no idea how I would use the ratio test on this sum. Also, the root test is inconclusive, because I'm just gonna get $\lim\limits_{n\to 0} \left(\frac{n}{n+1}\right)^{2}$, which is $1$.

(the limit symbol isn't showing up properly but I don't know how to do it better)

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    You get $\bigl(\frac{n}{n+1}\bigr)^n$ in the root test. – 2017-01-02
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    Yeah... pardon me for the language, but I don't know what the f**k I was thinking. – 2017-01-02

3 Answers 3

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It's a good idea to use the root test, but you simplified the expression incorrectly. Instead, you should get $$ \lim_{n\to\infty}\Big(\frac{n}{n+1}\Big)^{\frac{n^2}{n}}=\lim_{n\to\infty}\Big(\frac{n}{n+1}\Big)^{n}=\frac{1}{e}$$ and since $\frac{1}{e}<1$ it follows that the series converges.

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You misapplied the root test: you have to search the limit of $$\Bigl(\frac n{n+1}\Bigl)^n,$$ which is $1/\mathrm e$.

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I wouldn't recommend it, but the ratio test works as well:

$$\frac{\left(\frac{n+1}{n+2}\right)^{(n+1)^2}}{\left(\frac{n}{n+1}\right)^{n^2}} = \frac{\left(\frac{n+1}{n+2}\right)^{n^2}}{\left(\frac{n}{n+1}\right)^{n^2}}\cdot \left(\frac{n+1}{n+2}\right)^{2n+1} = \left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot \left(\frac{n+1}{n+2}\right)^{2n+1} = \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot\left(1-\frac 1{n+2}\right)^{2n+1}\longrightarrow e\cdot e^{-2}=\frac 1 e$$

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    Third downvote in couple of minutes for no good reason. Did I make someone cry? – 2017-01-03