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Finding whether the series $$\sum^{\infty}_{n=1}\frac{2}{n(1+\ln(n))^{3}}$$ is converges or Diverges.

I am trying to solve it using Integral test

Let $\displaystyle f(x)=\frac{2}{x(1+\ln(x))^3}=\frac{(1+\ln(x))^{-3}}{x}.$

Then $\displaystyle f'(x)=\frac{x\cdot -3(1+\ln(x))^{-4}\cdot \frac{1}{x}-(1+\ln(x))^{-3}}{x^2}<0$ for all $x\geq 1$

So function is decreasing function

Also $$\int^{\infty}_{1}\frac{1}{x(1+\ln(x))^{3}}dx$$

Put $1+\ln(x)=t$ and $\displaystyle \frac{1}{x}d=dt$ and changing limits

$$\int^{\infty}_{1}t^{-3}dt=-\frac{1}{t^2}\bigg|^{\infty}_{2}=0.25$$ which is finite no.

So the series is converges

What i have try is right or not. If not Then please tell me How to solve it . Thanks

jacky
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    It looks good! There's just a minor calculation error when you found the integral though. The antiderivative of $\cfrac{1}{t^{3}}$ is $\cfrac{-1}{2t^{2}}$, which should give you an answer of 0.5. – Hadi May 03 '20 at 19:50

1 Answers1

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The integral test or the Cauchy condensation test are perfectly viable ways to go. As an alternative you may notice that

$$ 2\left(\frac{1}{(1+\log(n))^2}-\frac{1}{(1+\log(n+1))^2}\right)=\frac{2(1+\log n+1+\log(n+1))\log\left(1+\frac{1}{n}\right)}{(1+\log(n))^2(1+\log(n+1))^2} $$ is greater than $\frac{2}{n(1+\log(n))^3}$ for any $n\geq 2$, so by creative telescoping

$$ \sum_{n\geq 1}\frac{2}{n(1+\log(n))^3}\leq 2+2\sum_{n\geq 2}\left(\frac{1}{(1+\log(n))^2}-\frac{1}{(1+\log(n+1))^2}\right) = 2+\frac{2}{(1+\log 2)^2}.$$

Jack D'Aurizio
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