8

Does

$$ \dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots $$

converge or diverge?

The following is my idea:

use $1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}>\ln{n}$

$$ \sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{(1+1/2+\cdots+1/n)} < \sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{\ln{n}} = \sum_{n=1}^{\infty}n^{-\ln(19/7)} $$ But $p=\ln{\dfrac{19}{7}}<1$, becasue $\dfrac{19}{7}\approx 2.71428<e=2.71828$

I guess the series is divergent because I use $$1+1/2+\cdots+1/n\approx \ln{n}, n\to\infty$$

to find $$\sum_{n=1}^{\infty} (1/x)^{1+1/2+1/3+\cdots+1/n}$$ is convergent only if $x>e$.

So, my question is: how do I determine whether

$$ \dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots $$

converges or diverges? Thank you

Rookatu
  • 1,752
math110
  • 93,304

2 Answers2

3

Since

$$ H_n = \sum_{k=1}^{n} \frac{1}{k} = \log n + \gamma + o(1), $$

we have, for $x > 0$,

$$ x^{H_n} \sim x^{\log n + \gamma} = x^\gamma n^{\log x} $$

As $n \to \infty$. We may then apply the limit comparison test to conclude that

$$ \sum_{n=1}^{\infty} \frac{1}{x^{H_n}} $$

converges when $\log x > 1 \Leftrightarrow x > e$ and diverges when $\log x \leq 1 \Leftrightarrow x \leq e$.

Since $19/7 < e$ the series in your question diverges.

Antonio Vargas
  • 24,993
1

It can be used Raabe's test.

We have series $$\sum_{n=1}^{\infty} a_n, \qquad \mathrm{where } \quad a_n = \left(\dfrac{7}{19}\right)^{1+\frac{1}{2}+\cdots+\frac{1}{n}}. $$

We wil construct value $R_n= n \left( \dfrac{a_{n+1}}{a_n}-1\right)$. Denote $R=\lim\limits_{n\to\infty} R_n$.

If $R<-1$, then series converges.
If $R>-1$, then series diverges.

$R_n = n\left(\left(\dfrac{7}{19}\right)^{\frac{1}{n+1}}-1\right)= n\left( \exp(\frac{1}{n+1}\ln\frac{7}{19})-1\right)$.

Using Taylor series for $\exp$, we get $$R_n = n \sum_{j=1}^{\infty} \dfrac{ \left(\frac{1}{n+1} \ln\frac{7}{19}\right)^j }{j!}.$$

So, $R=\lim\limits_{n\to\infty} R_n = \ln\frac{7}{19} >\ln \frac{1}{e}=-1$. (because $\frac{19}{7}<e$).

Actually, $R \approx -0,99852883011112715490367468844.\;$

So, series diverges.

Oleg567
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