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Find a formula for $\displaystyle\sum_{i=1}^n \frac{i}{(i+1)!}$ and prove that it holds for all $n \ge 1$.

How do you find an equation for this formula? Is it common sense or is there a way to find it? I'm not very good with sequences or series. Please show me how you would approach in tackling this problem. Thanks

user10444
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Roy Kesserwani
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    Hint: The general term is $\frac{(i+1)-1}{(i+1)!}$, which is $\frac{1}{i!}-\frac{1}{(i+1)!}$. Now sum. An awful lot of cancellation! Side comment: There is no "universal" method. – André Nicolas Apr 13 '14 at 21:47
  • One thing can of course help somewhat universally. Compute the sum explicitly for $n=1, 2, 3, 4$. The results are fractions - what do you notice about the relation between numerator and denominator? What do you notice about the dneominator (recognize the sequence)? Can you guess(!) the explcit formula from this? Can you then prove it by induction? – Hagen von Eitzen Apr 13 '14 at 21:55
  • @HagenvonEitzen, with such "recognize the sequence" games, you have to simplify "just enough" not to erase the pattern you are looking for. Try not simplifying at all, do simple modifications, ... Getting it right is an art, honed by trying lots of examples. – vonbrand Apr 13 '14 at 22:00
  • @AndréNicolas how do you prove 1/i! - 1/(i+1)!? – Roy Kesserwani Apr 14 '14 at 02:54
  • I wrote the proof, basically: $\frac{(i+1)-1}{(i+1)!}=\frac{i+1}{(i+1)!}-\frac{1}{(i+1)!}$. Finally, $\frac{i+1}{(i+1)!}=\frac{1}{i!}$. This is because $(i+1)!/(i+1)=i!$. – André Nicolas Apr 14 '14 at 02:59

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If you are interested in the limit as $n \to \infty$ of your sum you could use the fact that $\sum_{n=0}^{\infty}\frac{1}{(n+1)!}=e-1$. Denote by $s$ your sum. Then $$ s+e-1= 1+\sum_{i=1}^{\infty} \frac{i+1}{(i+1)!}=1+\sum_{i=1}^\infty \frac{1}{i!}=e $$ Therefore $s=1$.

Beni Bogosel
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