The given sequence is: $$ a_1 = 1; \quad a_{n+1} = \frac{2a_n}{n+1} $$ $n$ is natural. What is the limit if $n\to\infty$? Please help!
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2Have you tried working out the first few terms, and looking for a pattern? – Gerry Myerson Oct 10 '15 at 23:26
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Well it gets messy but it tends to zero. Unfortunately I need the steps of the proof, not just the initial (right) guess as an answer – Próba Oct 10 '15 at 23:58
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I find that, whenever there is an $n$ or $n+1$ in a recurrence, throwing a $n!$ or $(n+1)!$ somewhere usually helps.
You have $a_{n+1} = \frac{2a_n}{n+1} $.
If we multiply by $(n+1)!$, this becomes $(n+1)!a_{n+1} = 2n!a_n $. (Note that multiplying by $n!$ or dividing by $n!$ do not work.)
Letting $b_n =n!a_n $, we get $b_{n+1} =2b_n $, which you should be able to solve.
Then get $a_n$ by $\frac{b_n}{n!} $.
marty cohen
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This "throwing an n! somewhere" works a little too well here.... I'm suspicious. – ZKe Oct 11 '15 at 01:10
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Besides, if something works "too well", it is probably a homework problem. – marty cohen Oct 11 '15 at 02:23
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Once you have sorted out the closed form for $a_n$, you can try to determine convergence of yhe sum $\sum_n \frac{2^n}{n!}$. If it convrges, then the limit of the series is obvious.
Alex
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