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Does the following series converge or diverge? $\sum\limits_{j=1}^\infty \frac{(1+\frac{1}{j})^{2j}}{e^j}$

Using the ratio test, I found $|\frac{a_{j+1}}{a_j}|$ to be $|\frac{(1+\frac{1}{j+1})^{2j+2}}{e(1+\frac{1}{j})^{2j}}|$

To find the lim sup of this, I looked at the ratio as $j \rightarrow \infty$, and arrived at $\frac{1}{e}<1$, so by the ratio test, the series would converge absolutely.

Does this look valid? I wasn't sure about the step of taking the limit as $j \rightarrow \infty$ to find the lim sup.

Thank you!

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  • $\begingroup$ If you have the Root Test, it is very quick. The $j$-th root is $\frac{(1+1/j)^2}{e}$, which approaches $1/e\lt 1$. The Ratio Test calculation can also be pushed through. The one currently in the post has gaps. $\endgroup$ Nov 12 '15 at 6:50
  • $\begingroup$ About your proof, you did not show that the limit of the ratios is $1/e$. You reached an intuitive understanding that the limit should be $1/e$, but some work is needed for a proof. $\endgroup$ Nov 12 '15 at 7:12
  • $\begingroup$ Would the work still needed to close the gaps just be showing that as $j \rightarrow \infty$ the numerator becomes $1^{2j+2}$ and the denominator becomes $e(1)^{2j}$ and that ratio approaches $\frac{1}{e}$ as $j \rightarrow \infty$? $\endgroup$
    – byong
    Nov 12 '15 at 7:15
  • $\begingroup$ Although, I agree with you, I had not looked at the Root Test, but that makes the problem significantly faster $\endgroup$
    – byong
    Nov 12 '15 at 7:17
  • $\begingroup$ About your penultimate comment, you let the $j$ in $1+1/j$ go to infinity, getting $1$, and then let the exponent $2j$ go to infinity. That is wrong. The same way you could conclude that $(1+1/j)^j$ approaches $1$ while in fact it approaches $e$. $\endgroup$ Nov 12 '15 at 7:47
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You could also say this: the numerator goes to $e^2$ from below, so the series is less than $\sum \dfrac{e^2}{e^j}$, which converges by geometric test.

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  • $\begingroup$ That makes sense. Is the method that I used also correct though? $\endgroup$
    – byong
    Nov 12 '15 at 6:59
  • $\begingroup$ Yeah, your method is fine. Note that you can use limit to find the lim sup when the limit exists since, in such a case, the lim and lim sup are equal. $\endgroup$
    – MCT
    Nov 12 '15 at 7:18

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