Can you use either taking the limit or solve the integral to find out if $\displaystyle \sum_{n=1}^\infty\frac{n}{\sqrt{n^2+1}}$ diverges or converges?
Can you use either taking the limit or solve the integral to find out if $\displaystyle \sum_{n=1}^\infty\frac{n}{\sqrt{n^2+1}}$ diverges or converges?
the limit test for divergence is the easiest method.
$\displaystyle \lim_{n \to \infty} \frac {n}{ \sqrt {n^2 + 1}} \neq 0$ thus the series diverges
Can you use either taking the limit or solve the integral to find out if $\displaystyle \sum_{n=1}^\infty\frac{n}{\sqrt{n^2+1}}$ diverges or converges?
The problem becomes harder if we consider $\displaystyle \sum_{n=1}^\infty\frac{n}{\sqrt{n^3+1}}$ or $\displaystyle \sum_{n=1}^\infty\frac{n}{\sqrt{n^4+1}}$