Find the value of $\frac{\frac 12 . \frac 22}{1^3} + \frac{\frac 22 . \frac 32}{1^3 + 2^3} + \frac{\frac 32 \frac 42}{1^3 + 2^3 + 3^3}$ upto n terms

Question

For the numerator nth term will be $$\frac n2 . \frac {n+1}{2}$$ $$=\frac{n^2+n}{4}$$

For the denominator, nth term will be $$1^3 + 2^3 + 3^3 .....n^3$$ $$\frac{(n^2)(n+1)^2}{4}$$

So, the nth term of the whole series will be $$T_n=\frac{1}{(n)(n+1)}$$

So, it’s sum will be $$\sum T_n =?$$

I can’t solve further. What should I do now?

EDIT

AS Moustafa Ayaz correctly pointed out, the expression can be written as

$$\sum \frac 1n - \sum \frac {1}{n+1}$$

Had n been in th numerator, it would have been fairly easy to solve, but because it’s not, I am having trouble in simplifying It. What should I do next?

Answer 1

Hint $${1\over n{(n+1)}}={1\over n}-{1\over n+1}$$