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$$S=\sum_{k=1}^\infty\left(\frac{k}{k+1}\right)^{k^2};\hspace{10pt}S_n=\sum_{k=1}^n\left(\frac{k}{k+1}\right)^{k^2}$$ Let $S_n$ represent its partial sums and let $S$ represent its value. Prove that $S$ is finite and find an n so large that $S_n$ approximate $S$ to three decimal places.

Solution: first of all, I think that we Will use L'hopital rule and then use root test while starting to solve this. But how?

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Hint: use the $k$th root test.


If $a_k$ are the coefficients, that is:

$$a_k=\left(\frac{k}{k+1}\right)^{k^2}$$

then:

$$\sqrt[k]{a_k}=\left(\frac{k}{k+1}\right)^{k}=\left(1-\frac{1}{k+1}\right)^k$$

With some fantasy, do you recognize this sequence? Does it converge to something you know?

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I prefer the following: $a_k^{1/k} = \left ( 1 + \frac{1}{k} \right )^{-k}$. – Ron Gordon Dec 22 '12 at 15:07
    
@rlgordonma: I didn't want to completely give it away, but yes, that is closer to an answer :). – akkkk Dec 22 '12 at 15:08

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