Determine if this series : $\sum_{n=1}^\infty \frac{n}{(n+1)\ln(n+1)}$ converges or diverges.

Question

I have this infinite series: $$\sum_{n=1}^\infty \frac{n}{(n+1)\ln(n+1)}$$ I am really confused on where to begin with, any hint would be appreciated.

Answer 1

Note that

$$\frac{n}{(n+1)ln(n+1)} \sim \frac1{\ln n}$$

therefore the given series diverges by limit comparison test with $\sum \frac1{\ln n}$ since

$$\frac1{\ln n}>\frac1n$$

and then $\sum \frac1{\ln n}$ diverges by comparison with $\sum \frac1{n}$.

Answer 2

$\lim_n{n\over{ln(n+1)}}=+\infty$ and $\sum{1\over{n+1}}$ diverges so the serie diverges.

You can also says that ${{n\over {(n+1)\ln(n+1)}}\over {1\over {n+1}}}$ $={n\over{\ln(n+1)}}$ and $\lim_n{n\over{\ln(n+1)}}=+\infty$ the comparison test implies that the series diverges.

Answer 3

From $\ln(n+1)<n+1$ and $n>\dfrac{n+1}{2}$ then $$\sum_{n=1}^\infty \frac{n}{(n+1)\ln(n+1)}>\dfrac12\sum_{n=1}^\infty \frac{1}{n+1}$$

Answer 4

$a_n=\dfrac{n}{(n+1)\ln(n+1)} > \dfrac{1}{(n+1)\ln(n+1)}\implies \displaystyle \sum_{n=1}^\infty a_n \ge\displaystyle \sum_{n=2}^\infty\dfrac{1}{n\ln n}$. Now use the Cauchy Condensation Test to the right series we have: $\displaystyle \sum_{n=2}^\infty 2^n\cdot \dfrac{1}{2^n\ln (2^n)}= \dfrac{1}{\ln 2}\displaystyle \sum_{n=2}^\infty \dfrac{1}{n} = \infty$. This shows the former series diverges.... Note: You can solve it by first use comparison and integral test.