I'm working through Spivak's Calculus (Fourth Edition).

In chapter 2, problem 6:

The formula for $1^2 + ... + n^2$ may be derived as follows. We begin with the formula $(k+1)^3 - k^3 = 3k^2 +3k +1$
Writing this formula for $k = 1, ... , n$ and adding, we obtain
$(n+1)^3 - 1= 3 (1^2 + ... + n^2 ) +3(1+...+n) +n$
Thus we can find $(1^2 + ... + n^2 )$ if we already know $(1+...+n)$ (which could have been found in a similar way).

This is clear for me. However, I'm stuck in this problem: Use this method to find $$\frac{3}{1^2\times2^2} + \frac{5}{2^2\times3^2} +\cdots+ \frac{(2n+1)}{n^2(n+1)^2}$$

I would like to know how could the one think in this and not just how to solve it. I want to know how to deal with "deriving summation formulas" in general.

  • $\begingroup$ i don't think that this method will work here $\endgroup$ Jan 29, 2017 at 13:11
  • $\begingroup$ @Dr.SonnhardGraubner It worked for $1/n(n+1)$ In a way that may be complex for me. So I think it's possible to do it the same way for this expression. $\endgroup$
    – Ahmed99
    Jan 29, 2017 at 13:13
  • $\begingroup$ Do you mean $3/(1^2\cdot 2^2)$ and so on? $\endgroup$
    – Wolfram
    Jan 29, 2017 at 13:13
  • $\begingroup$ @Wolfram Yeah, could you edit the question to make it looks clearer. $\endgroup$
    – Ahmed99
    Jan 29, 2017 at 13:14

1 Answer 1


I think what was meant here is that $\frac{2k+1}{k^2(k+1)^2}=\frac1{k^2}-\frac1{(k+1)^2}$. You can use this formula to replace each member of the sum by the difference and then cancel out similar terms.

  • 1
    $\begingroup$ Yes. I got the formula for the summation: $1-1/(n+1)^2$. $\endgroup$
    – Ahmed99
    Jan 29, 2017 at 13:28
  • $\begingroup$ @Ahmed99 Then you should be good? $\endgroup$ Jan 29, 2017 at 13:35
  • $\begingroup$ @SimplyBeautifulArt Yeah. Asking about how could the one deal with such problems in general, may not be appropriate for the site. But I asked this because I spent almost an hour staring in such a simple question. $\endgroup$
    – Ahmed99
    Jan 29, 2017 at 13:45
  • $\begingroup$ @Ahmed99 I'm afraid, there is no general method to search closed forms of different sums. There are only a bunch of tricks, and the one shown here is trying to write the term of the sum $f(n)$ as the difference $g(n+1)-g(n)$ for some function $g$. Then the sum $f(1)+\cdots+f(n)$ is obviously $g(n+1)-g(1)$. But it is neither easy to find $g$ in the general case, nor it always works. $\endgroup$
    – Wolfram
    Jan 29, 2017 at 13:51
  • $\begingroup$ @Wolfram I see. Thanks for your clarification. $\endgroup$
    – Ahmed99
    Jan 29, 2017 at 15:57

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