Please share some good questions of Calculus for JEE Advanced.

# Thread to share Good Questions from Calculus which requires deep thinking

Havent solved it...but i think substituting x=1/t can help.

@ananda_krishnan

Will look at it afterwards @ananda_krishnan

After i try it out.

@Unik-Wadhwani It is from brilliant.Take a look at solution:

I loved it ..so shared it!

As yu wish @Unik-Wadhwani

Let \displaystyle \mathcal{S} = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)} = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{n!}{n^kk!(n-k)!(k+3)}

\displaystyle = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{n(n-1)(n-2)\cdots (n-(k-1))}{n^k k!(k+3)}

\displaystyle =\sum^{\infty}_{k=0}\frac{1}{k!(k+3)} = \sum^{n}_{k=0}\frac{(k+1)(k+2)}{(k+3)!}

\displaystyle = \sum^{\infty}_{k=0}\frac{k(k+3)+2}{(k+3)!} = \sum^{\infty}_{k=0}\frac{(k+2)-2}{(k+2)!}+2\sum^{\infty}_{k=0}\frac{1}{(k+3)!}

\displaystyle =\sum^{\infty}_{k=0}\frac{1}{(k+1)!}-2\sum^{\infty}_{k=0}\frac{1}{(k+2)!}+2\sum^{\infty}_{k=0}\frac{1}{(k+3)!}

\displaystyle = (e-1)-2(e-1-1)+2\bigg(e-1-\frac{1}{2}\bigg)

So we get \displaystyle \mathcal{S} = e-2.