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?

I understood it using the last answer on the page. But can I prove it without using numbers and instead just use variable to form a general proof? You may have to look the answer provided to understand what I am saying. — Aditya, Sep 18 '19 at 14:50
No it's not clear to me what you are asking. If you're asking for an alternative to partial fractions, then the above link has such an answer, but why wouldn't you want to use partial fractions? — Simply Beautiful Art, Sep 18 '19 at 14:53
I am talking in context of the last answer provided to the question. The guy has put in values in the partial fraction to prove his point, which is perfectly fine, but can it be done by using a general proof in the form of variables? Looking at that is answer is really important to understand what I am saying, — Aditya, Sep 18 '19 at 14:56
Which answer is last depends on how you are sorting posts. And there are multiple answers there. If you understand them, you would be able to see why they work and that they do not depend on plugging in values. — Simply Beautiful Art, Sep 18 '19 at 14:59

score 5 · Answer 1 · answered Sep 18 '19 at 14:31

5

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

answered Sep 18 '19 at 14:31

Mostafa Ayaz

31,924

Is there a way to do this that does not involve splitting the denominator at all? – Aditya Sep 18 '19 at 14:34
1

It's called "partial fraction expansion" and it is very useful in all sorts of situations: why would you not want to use it? – NickD Sep 18 '19 at 14:37
Cause I am not able to solve summations like that. – Aditya Sep 18 '19 at 14:39
I'm afraid that there are no other ways – MafPrivate Sep 18 '19 at 14:40
Note that $$\sum{1\over n}-{1\over n+1}\ne \sum{1\over n}-\sum{1\over n+1}$$when the upper bound is infinite. When being finite, just simply delete the similar terms with opposite signs from both the $\sum$s – Mostafa Ayaz Sep 18 '19 at 14:55
No idea about what you said – Aditya Sep 18 '19 at 15:04

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

1 Answers1