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Check the convergence of the series $$ \sum_{n=1}^{\infty} \left(1+\frac{1}{\sqrt n}\right)^{-n^{3/2}}$$

I checked this equation with the test for divergence but it came out 0. So, now I don't know the nature of the following equation. Root test and ratio test and limit test are not going to work (or maybe I am doing it wrong) and I don't know how to use comparison test in this question.

Besides, I am having problem with the power i.e $-n^{\frac{3}{2}}$ in the following question, how to remove it?

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    $\begingroup$ Do you mean $$\sum_{n=1}^{\infty} (1+\frac{1}{\sqrt n})^{-n^{3/2}}?$$ $\endgroup$ – Professor Vector Sep 2 '17 at 14:20
  • $\begingroup$ Please check again your exponent 3/2. $\endgroup$ – hamam_Abdallah Sep 2 '17 at 14:35
  • $\begingroup$ @ProfessorVector yes $\endgroup$ – Iti Shree Sep 2 '17 at 14:43
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Apply the root test, then $$\lim_{n\to\infty} \sqrt[n]{|a_n|}=\lim_{n\to\infty}\left(1+\frac{1}{\sqrt n}\right)^{-\frac{n^{\frac{3}{2}}}{n}}=\lim_{n\to\infty}\left(1+\frac{1}{\sqrt n}\right)^{-\sqrt{n}}=e^{-1}$$ where we used the known limit $$\lim_{x\to +\infty}\left(1+\frac{1}{x}\right)^x=e.$$ Since the limit $e^{-1}$ is less than $1$, the series is convergent.

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  • $\begingroup$ I had just one doubt, after using root test we can use another test to proceed as well? $\endgroup$ – Iti Shree Sep 2 '17 at 17:24
  • $\begingroup$ @Iti Shree If the limit of $\sqrt[n]{|a_n|}$ is less than $1$ then root-test criterion says that the series is convergent. You don't need another test. $\endgroup$ – Robert Z Sep 2 '17 at 17:29
  • $\begingroup$ I wasn't asking for this question but for some cases, if I stuck at a question. $\endgroup$ – Iti Shree Sep 4 '17 at 7:57
  • $\begingroup$ @Iti Shree Sorry for that. So what is your question? $\endgroup$ – Robert Z Sep 4 '17 at 8:25
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To check the convergence of the series, one can forget the root test or equivalents or whatnots, and work directly with "hard" inequalities, since, for every $x$ in $[0,1]$, $$1+x\geqslant2^x\tag{$\ast$}$$ hence, for every $n\geqslant1$, $$\left(1+\frac1{\sqrt{n}}\right)^{-n\sqrt{n}}\leqslant\left(2^{1/\sqrt{n}}\right)^{-n\sqrt{n}}=2^{-n}$$ Thus, the series converges (and its sum is less than $1$).


To show $(\ast)$, note that the function $u:x\mapsto2^x$ is convex, that $y=1+x$ is the equation of the chord of the graph of $u$ between the points of affixes $x=0$ and $x=1$, and that $(\ast)$ simply states that the graph of $u$ is below its chord on $[0,1]$.

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The $n$th term is

$$\tag 1 \left (\frac{1}{(1+1/\sqrt n)^{\sqrt n}}\right)^n.$$

Now $(1+1/\sqrt n)^{\sqrt n} \to e >2,$ hence $(1)\le (1/2)^n$ for large $n.$ Thus the series converges by the comparison test.

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By limit comparison test

and using the fact that when $n\to +\infty $,

$$-n^\frac 32\ln (1+\frac {1}{\sqrt {n}})=$$ $$-n^\frac 32(\frac {1}{\sqrt {n}}-\frac {1}{2n}+\frac{1}{3n\sqrt {n}}-\frac {1}{4n^2}(1+\epsilon (n)) $$

thus $$u_n\sim e^{-n+\frac {\sqrt {n}}{2}-\frac {1}{3}+\frac {1}{4\sqrt {n}} }$$

the series converges.

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  • $\begingroup$ math.oregonstate.edu/home/programs/undergrad/… according to this the result after limit comparison test will be $\infty$ which is violating the rule, then how did you proceed? $\endgroup$ – Iti Shree Sep 2 '17 at 14:47
  • $\begingroup$ This is wrong, you should consider $$u_n=e^{-n^{\frac 32}\ln (1+\frac {1}{\sqrt {n}})},$$ instead. $\endgroup$ – Professor Vector Sep 2 '17 at 14:48
  • $\begingroup$ @ProfessorVector is it good now. thnx. $\endgroup$ – hamam_Abdallah Sep 2 '17 at 14:54
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    $\begingroup$ The equivalent of the logarithm does not imply the equivalent $u_n\sim e^{-n}$. Recall that $$u_n\sim v_n$$ means, if $v_n\ne0$ for every $n$, that $$\lim\frac{u_n}{v_n}=1$$ $\endgroup$ – Did Sep 2 '17 at 17:29
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    $\begingroup$ But to avoid all that fun, note that writing $u_n = e^{-n+o(n)}$ is enough to conclude. $\endgroup$ – Clement C. Sep 3 '17 at 14:48

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