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Tell, if the series converges.

Please verify my answer.

$$\displaystyle \sum_{n=2}^\infty\dfrac{\exp(n)}{\exp(n\sqrt[n]{n})\ln^{2}n}=\sum_{n=2}^\infty\dfrac{\exp(n)}{\exp(n)\frac{1}{n}\exp(n)\ln^{2}n}=\sum_{n=2}^\infty\dfrac{n}{\exp(n)\ln^{2}n}$$

Comparison test

$$\displaystyle \lim_{n\to \infty}\frac{n+1}{\exp(n+1)\ln^{2}(n+1)}\frac{\exp(n)\ln^{2}n}{n}=\underset{n\rightarrow\infty}{\lim}\frac{1}{e}\frac{n+1}{n}\frac{\ln^{2}n}{\ln^{2}(n+1)}=\frac{1}{e}$$

The series converges.

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You seem to have put $$e^{n\sqrt[n]n}=e^n\frac{1}{n}e^n$$ This doesn't look right. Perhaps you meant $$e^{n\sqrt[n]n}=e^{n^{n+1/n}}\;\;?$$ – DonAntonio Jan 9 '13 at 13:26
That's not true: $exp(n\sqrt[n]{n}) = exp(n) * exp(\sqrt[n]{n})$ ? – Joggi Jan 9 '13 at 13:28
I'm sorry @Joggi, but I'm afraid you'll have to review your notes in exponents theory. It must be $$a^{x+y}=a^x\cdot a^y\;\;,\;\;a^{xy}=(a^x)^y\Longrightarrow e^{n\sqrt[n]n}=(e^n)^{\sqrt[n]n}=(e^{\sqrt[n]n})^n\neq e^n\cdot e^{\sqrt[n]n}$$ – DonAntonio Jan 9 '13 at 13:33
No! I think, @Joggi, you're confusing big time the exponential function with the logarithmic function. With the last one we certainly have $$\log\sqrt[n]n=\frac{1}{n}\log n$$but not with its inverse function... – DonAntonio Jan 9 '13 at 13:53
No, @Joggi: $\,\log n^2=2\log n\neq \log^2n=(\log n)^2\,$ I honestly think you should read the basic of exponential and logarithmic functions before engaging in calculus any further. When dealing with series, one usually expect student to already know all thi stuff from high school! So you forgot, no big deal...but then try to fix this ASAP. – DonAntonio Jan 9 '13 at 14:08

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