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Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2). Not sure when to apply the induction hypothesis and how to get the result $1- \frac{1}{(n+2)!}$. Please help! thanks guys, youre the greatest!

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4 Answers 4

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Hint: $$1-\frac1{(n+1)!}+\frac{n+1}{(n+2)!} = 1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!} = 1-\frac{1}{(n+2)!}.$$

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  • $\begingroup$ That is more than a hint $\endgroup$
    – Henry
    Oct 10, 2013 at 21:05
  • $\begingroup$ I seems that I have a very hard time giving right hints. My hints either are called "more than a hint" or someone else gives more detailed answer after me and gets much more upvotes than me. :-) $\endgroup$
    – njguliyev
    Oct 10, 2013 at 21:11
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    $\begingroup$ I was not objecting to your answer (which I upvoted). Just the word "Hint" $\endgroup$
    – Henry
    Oct 10, 2013 at 21:14
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Induction assumption assumes your statement holds for $n=k$ Step after induction assumption. Let $n=k+1$ then the right hand side of your statement is, $$ 1-\cfrac{1}{(k+1)!}+\cfrac{k+1}{(k+2)!}.$$

Take common denominator of last two terms and you get $$ 1+\cfrac{-k+-2+k+1}{(K+2)!}=1-\cfrac{1}{(k+2)!}$$

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  • $\begingroup$ Wow, im an idiot and i forgot to add the (k+1)/(k+2)! to the left side, thanks for waking my brain up guys, much appreciated! $\endgroup$
    – user99714
    Oct 10, 2013 at 21:11
  • $\begingroup$ you are welcome :) $\endgroup$
    – triomphe
    Oct 10, 2013 at 21:12
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$$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$$

Note that this idea can also be used to make the sum into a telescopic sum.

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You can also prove this by power series manipulation (generating functions). Note first that changing the sum by starting at $i=0$ instead of $i=1$ doesn't change its value. Compute as follows:

\begin{align}\sum_{n=0}^\infty \sum_{i=1}^n \frac{i}{(i+1)!}x^{n+1} &= \sum_{n=0}^\infty \sum_{i=0}^n \frac{i}{(i+1)!}x^{n+1} \\\\&= \sum_{i=0}^\infty \sum_{n=i}^\infty \frac{i}{(i+1)!}x^{n+1} \\\\ &= \sum_{i=0}^\infty \frac{i}{(i+1)!} \left( \sum_{n=-1}^\infty x^{n+1} - \sum_{n=-1}^{i-1} x^{n+1}\right) \\\\&= \sum_{i=0}^\infty \frac{i}{(i+1)!} \left(\frac{1}{1-x}-\frac{1-x^{i+1}}{1-x} \right) \\\\&= \sum_{i=0}^\infty \frac{i}{(i+1)!}\frac{x^{i+1}}{1-x} \\\\&= \frac{x^2}{1-x} \sum_{i=0}^\infty \frac{ix^{i-1}}{(i+1)!} \\\\&= \frac{x^2}{1-x} \frac{d}{dx} \sum_{i=0}^\infty \frac{x^i}{(i+1)!} \\\\&= \frac{x^2}{1-x} \frac{d}{dx} \frac{e^x-1}{x} \\\\&= \frac{x^2}{1-x} \frac{xe^x-(e^x-1)}{x^2} \\\\&=\frac{1}{1-x}-e^x \\\\&=\sum_{n=0}^\infty x^n - \sum_{n=0}^\infty \frac{x^n}{n!} \\\\&= \sum_{n=0}^\infty \left(1-\frac{1}{n!}\right)x^n \\\\&= \sum_{n=1}^\infty \left(1-\frac{1}{n!}\right)x^n \\\\&= \sum_{n=0}^\infty \left(1-\frac{1}{(n+1)!}\right)x^{n+1} \end{align} Coefficients of like powers of $x$ in two equal power series must be equal, proving that $$\sum_{i=1}^n \frac{i}{(i+1)!} = 1-\frac{1}{(n+1)!},$$ as desired.

This can all be justified either as formal power series manipulation or as calculations with absolutely convergent series for $|x| < 1.$

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