Is the series $\sum^\infty_{n=1}\left( a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2 \right)$ converging?

Question

I've been trying to solve this for limit comparison test with $a_n=a^\frac{1}{n}+a^{-\frac{1}{n}}-2 , b_n= \frac{1}{n}$, but $\frac{a_n}{b_n}\rightarrow\ln{a}(a^{\frac{1}{x}}-a^{-\frac{1}{x}})\rightarrow 0$. Any help appreciated.

Use that $a^x=e^{x\ln a}=1+x\ln a+(x\ln a)^2/2+$ up to higher order terms for small $x$, with $x=1/n$. For comparison $b_n=1/n^2$ and this series converges. — Conifold, Jan 21 '20 at 03:47
It might help to note that $a^{2r}+a^{-2r}-2=(a^r-a^{-r})^2$. — Gerry Myerson, Jan 21 '20 at 05:08
@Conifold Thank you very much. But my teacher didn't teach $e^{x\ln{a}}=1+x\ln{a}+(x\ln{a})^2/2+...$. — 余志祥, Jan 21 '20 at 05:38
@GerryMyerson Thank you. But I didn't know how to apply this hint to solve. Could you give me more steps? — 余志祥, Jan 21 '20 at 05:40
User Pythagoras used the idea in my comment as the first line in his post, so no need now for me to elaborate. If you are happy with the answer Pythagoras posted, let me encourage you to "accept" it by clicking in the check mark next to it. — Gerry Myerson, Jan 21 '20 at 17:38
The expansion given by @Conifold is just the trivial Taylor expansion for $\exp$. How can your teacher not teach you this? Please look at this post and the linked posts for more on this standard technique. Any other solutions are ad-hoc and can often be found only by using asymptotic expansions in the first place. If you have trouble with asymptotic expansion, you can ask me in this chat-room. — user21820, Feb 09 '20 at 05:34

Pythagoras · Accepted Answer · 2020-01-21T05:52:57.820

1

Assume $a>0$. Let $x_n=\frac 1{2n}.$ Then $$\sum_{n=1}^\infty \left(a^{1/n}+a^{-1/n}-2\right)=\sum_{n=1}^\infty(a^{x_n}-a^{-x_n})^2$$ $$=\sum_{n=1}^\infty a^{2x_n}(1-a^{-2x_n})^2=\sum_{n=1}^\infty a^{2x_n}(1-e^{-2x_n\ln a})^2,$$ which by mean value theorem equals $$\sum_{n=1}^\infty a^{2x_n}(e^{-2y_n\ln a}\cdot 2x_n\ln a)^2,$$ where $0<y_n<x_n$. Clearly the series is bounded by $$C\cdot \sum_{n=1}^\infty (2x_n)^2=C\sum_{n=1}^\infty\frac 1{n^2}$$ where $C$ is a bounded constant. It follows that the original series is convergent.

Note: $1-e^x=e^0-e^x=e^{\xi}(0-x),$ where $\xi$ is between $0$ and $x$.

edited Jan 21 '20 at 05:52

answered Jan 21 '20 at 05:44

Pythagoras

7,079

Is it by Dirichlet comparason test? – 余志祥 Jan 21 '20 at 06:16
No, just the usual one. – Pythagoras Jan 21 '20 at 06:21
$C$ is some bounded number times $(\ln a)^2$, as $x_n$ and $y_n$ both approach $0$. – Pythagoras Jan 21 '20 at 06:29
1

I think I got it. Thanks for your help. – 余志祥 Jan 21 '20 at 06:50

score 0 · Answer 2 · answered Jan 21 '20 at 06:02

Let be $a>0$. Since $$ a^x + a^{ - x} - 2 = 2\left[ {\cosh (x\ln a) - 1} \right] $$ and since $$ \mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\cosh (t) - 1}} {{t^2 }}} \right] = \frac{(\ln a)^2} {2} $$ by setting $ t = \frac{{\ln a}} {n} $ you get $$ \begin{gathered} \mathop {\lim }\limits_{n \to + \infty } n^{ - 2} \left( {a^{\frac{{\text{1}}} {n}} + a^{ - \frac{{\text{1}}} {n}} - 2} \right) = \hfill \\ = \mathop {\lim }\limits_{n \to + \infty } 2n^{ - 2} \left[ {\cosh \left( {\frac{{\ln a}} {n}} \right) - 1} \right] = \frac{(\ln a)^2} {2} \hfill \\ \hfill \\ \end{gathered} $$ and your series is convergent by asymptotic comparison test.

Is the series $\sum^\infty_{n=1}\left( a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2 \right)$ converging?

2 Answers2