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Given that for a sequence $\{a_n\}$ where $1 \leq a_n \leq n-1$ is it true that $\sum_{n=2}^{\infty}\frac{a_n}{n!}=1$ implies $a_n=n-1 \forall n \geq 2$ .How to prove it ?

This is a sufficient condition which is very obvious. But whether it is necessary that is the question.

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I am not very much comfortable with analysis and all.. but, It would be helpful if you can write what are your thoughts.... –  Praphulla Koushik Oct 3 '13 at 6:52

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HINT: Rewrite $\frac{n-1}{n!}$ as $\frac{n}{n!}-\frac1{n!}$ and simplify to show that $\sum_{n\ge 2}\frac{n-1}{n!}=1$, and observe that decreasing one or more terms decreases the sum of the series.

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M.Scott: Why will it decrease ? If I decrease one term of an infinite series which converges, will the sum decrease ? I am not sure –  sosha Oct 3 '13 at 8:07
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@prasenjit: Yes, it will. Suppose that $\sum_{n\ge 0}a_n=L$, and I decrease $a_{10}$ to $a_{10}-c$. Let the original partial sums be $s_n=\sum_{k=0}^na_k$, and let the new partial sums be $s_n'$. Then $s_n'=s_n$ when $n<10$, but $s_n'=s_n-c$ for all $n\ge 10$, so the sum of the new series is $\lim_{n\to\infty}s_n'=L-c$. –  Brian M. Scott Oct 3 '13 at 8:11
    
To prasenjit: A finite number of terms don't affect convergence/divergence of a series, but their values do affect the sum of the series in case it is a convergent one. –  Paramanand Singh Oct 4 '13 at 4:39

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