Sum......

Seems like I have seen this question on TIIT only, some before.

We can first prove that

\frac{1}{(k+1)\sqrt k} < 2 \left(\frac{1}{\sqrt k} - \frac{1}{\sqrt{k+1}} \right).

i.e. that \frac{1}{\sqrt{k+1}} < \frac{2}{\sqrt k + \sqrt{k+1}} \right) or that \sqrt{k+1} > \sqrt k which is obvious.

Hence the sum can be bounded by

2 \left(1 - \frac{1}{10} \right)<2

For k≥4, we have

\frac{1}{(k+1)\sqrt k} \ge \frac{2}{(k+1)k} = 2 \left(\frac{1}{k} - \frac{1}{k+1} \right)

Hence the given sum is greater than

\frac{1}{2} + \frac{1}{3 \sqrt 2} + 2 \left(\frac{1}{4} - \frac{1}{100} \right)>\frac{1}{2} + \frac{1}{2} + \left(\frac{1}{9} - \frac{1}{50} \right)>1

Hence [A]=1

Great Sir

Thanks hsbhatt Sir