Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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.

share|cite|improve this answer
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 – RIchard Williams Oct 3 '13 at 8:07
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.