The series \(\frac {1}{2^2 + 1} + \frac {\sqrt 2}{3^2 + 1} + \frac {\sqrt 3}{4^2 + 1} + ...\) is-

Concept:

If the given is infinite series, find the partial sum (i.e. sum of first n^th term )

If the sequence of partial sums is a convergent sequence (i.e. its limit exists and is finite) then the series is also called convergent 

Calculations:

Consider the  given series is  \(\frac {1}{2^2 + 1} + \frac {\sqrt 2}{3^2 + 1} + \frac {\sqrt 3}{4^2 + 1} + ...\)

The n^th term of the series is a_n = \(\rm\dfrac{\sqrt n}{(n+1)^2+1}\)

Given is infinite series, find the partial sum (i.e. sum of first nth term )

If the sequence of partial sums is a convergent sequence (i.e. its limit exists and is finite) then the series is also called convergent 

⇒ an = \(\rm\dfrac{\sqrt n}{(n+1)^2+1}\)

⇒\(\rm\\lim_{n \to \infty }a_n=\lim_{n \to \infty }\frac{\sqrt n}{(n+1)^2+1}\)

 ⇒\(\rm\\lim_{n \to \infty }a_n=\lim_{n \to \infty }\frac{\sqrt n}{n^2+2n+2}\)

⇒\(\rm\\lim_{n \to \infty }a_n=\lim_{n \to \infty }\frac{\sqrt n}{\sqrt n(n^{\frac3 2}+2\sqrt n+\frac{2}{\sqrt n})}\)

⇒\(\rm\\lim_{n \to \infty }a_n=0\), limit exists and is finite

The series \(\frac {1}{2^2 + 1} + \frac {\sqrt 2}{3^2 + 1} + \frac {\sqrt 3}{4^2 + 1} + ...\) is convergent.