Question

Find the sum of infinite series - 3/4+5/36+7/144+9/400+11/900+.......

Answer 1

\[\begin{align*}\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+\cdots&=\frac{3}{2^2}+\frac{5}{6^2}+\frac{7}{12^2}+\frac{9}{20^2}+\cdots\\
&=\frac{3}{(2\times1)^2}+\frac{5}{(3\times2)^2}+\frac{7}{(4\times3)^2}+\frac{9}{(5\times4)^2}+\cdots\\
&=\sum_{n=1}^\infty\frac{2n+1}{(n(n+1))^2}
\end{align*}\]
By the way, if you ever have trouble figuring out a pattern in a given sequence, give https://oeis.org/ a try.

Anyway, what you can do here is split the expression into partial fractions:
\[\begin{align*}\frac{2n+1}{n^2(n+1)^2}&=\frac{a}{n}+\frac{b}{n^2}+\frac{c}{n+1}+\frac{d}{(n+1)^2}\\
2n+1&=an(n+1)^2+b(n+1)^2+cn^2(n+1)+dn^2\\
2n+1&=an^3+2an^2+an+bn^2+2bn+b+cn^3+cn^2+dn^2\\
2n+1&=(a+c)n^3+(2a+b+c+d)n^2+(a+2b)n+b
\end{align*}\]
So you have the system
\[\begin{cases}a+c=0\\2a+b+c+d=0\\a+2b=2\\b=1\end{cases}~~\Rightarrow~~a=0,~b=1,~c=0,~d=-1\]
So,
\[\sum_{n=1}^\infty\frac{2n+1}{(n(n+1))^2}=\sum_{n=1}^\infty\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\]
which is a telescoping series.

Answer 2

just an observation - the partial fractions step is general and always works ! but if you're lucky, sometimes u can avoid the work by guessing :

\[2n+1 = (n+1)^2 - n^2\]

Answer 3

thats for @Jhimli :)

Answer 4

@ganeshie8 I'm a fan of the tedious method :P

Answer 5

haha me too lol, the generic(tedius) method is very effeictive for teaching :D

Answer 6

Thanks a lot! :-)