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I am trying to determine the convergence of the series below: $$\sum_{n=1}^{\infty}{\frac{(n+1)}{(n^2+2)\ln(n+3)}}$$

I've tried comparison test, Cauchy-condesation, but nothing seems to work.

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    $\begingroup$ Compare it with math.stackexchange.com/q/1522832/42969 $\endgroup$
    – Martin R
    May 22, 2022 at 6:41
  • $\begingroup$ Applying Cauchy-condensation gives me $ln(2^n+3)$ and I can not get the n out of the log $\endgroup$
    – Kain
    May 22, 2022 at 6:50
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    $\begingroup$ A general strategy is to compare with an easier series, such as the one Martin mentions. $\endgroup$
    – Kenta S
    May 22, 2022 at 8:17

2 Answers 2

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$$ \sum_{n=1}^{\infty}{\frac{n+1}{(n^2+2)\ln(n+3)}} \ge \sum_{n=3}^{\infty}{\frac{n}{2\cdot n^2 \cdot\ln(2n)}} = \frac{1}{2}\sum_{n=3}^{\infty}{\frac{1}{ n\ln(2n)}} $$ Now, by the Cauchy's condensation test we have that $$ \sum_{n=3}^{\infty}{\frac{1}{ n\ln(2n)}} $$ converges iff $$ \sum_{n=3}^{\infty}{\frac{2^n}{ 2^n\ln(2\cdot 2^{n})}} $$ converges. But $$ \sum_{n=3}^{\infty}{\frac{2^n}{ 2^n\ln(2\cdot2^{n})}} =\sum_{n=3}^{\infty}{\frac{1}{ (n+1)\ln(2)}} = \frac{1}{\ln(2)}\cdot\sum_{n=3}^{\infty}{\frac{1}{ n+1}} = +\infty $$ by the divergence of the harmonic series. It follows that $$ \sum_{n=1}^{\infty}{\frac{n+1}{(n^2+2)\ln(n+3)}} \ge \frac{1}{2}\sum_{n=3}^{\infty}{\frac{1}{ n\ln(2n)}} = +\infty. $$

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  • $\begingroup$ Thanks, I have a question, could I also use the integral test to show that $\sum_{n=3}^{\infty}{\frac{1}{nln(2n)}} $ diverges ? $\endgroup$
    – Kain
    May 22, 2022 at 7:00
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    $\begingroup$ @Kain yes you can. The integral corresponding to that series is $$\int^{\infty}_3\frac{1}{n\ln(2n)}\text{ d}n$$ simply do a $u$ sub with $u=\ln(2n)$ and the integrand becomes $\frac1u$, which will diverge. $\endgroup$
    – Max0815
    May 22, 2022 at 7:08
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    $\begingroup$ Yes. Then to prove that $\int_3^\infty \frac{1}{x \ln(2x)} \mathrm{d}x$ you could make the change of variables $x = e^u$. $\endgroup$
    – Bob
    May 22, 2022 at 7:08
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Using the integral test, it is trivial to show that $$\sum^{\infty}_{n=2}\frac{1}{n\ln n}$$ diverges. We may match indices with this from your series by doing the following $$\sum_{n=1}^{\infty}{\frac{(n+1)}{(n^2+2)\ln(n+3)}}=\frac{2}{3\ln4}+\sum_{n=2}^{\infty}{\frac{(n+1)}{(n^2+2)\ln(n+3)}} $$ By taking the limit comparison test, we can show that the behavior of the two summands $\dfrac{1}{n\ln n}$ and $\dfrac{n+1}{(n^2+1)\ln(n+1)}$ have the exact same behavior: the rational powers on the latter cancel to match the behavior of $\frac1n$, while the constant in the logarithm vanishes at infinity to match the behavior of $\frac1{\ln n}$.

Thus $$ \sum_{n=1}^{\infty}{\frac{(n+1)}{(n^2+2)\ln(n+3)}} $$ diverges

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    $\begingroup$ This is also a good solution thanks !! $\endgroup$
    – Kain
    May 22, 2022 at 7:08

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