Question

Consider a_n and the sum \[\sum_{n=1}^{\infty}a_n \] a_n is defined as \[a_{n+1}=\frac{a_n-a_{n+1}}{n}, a_1=1\]. Does the series converge or diverge?
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My steps:
Isolated for a_n to get:
\[a_n=(a_{n+1})n+(a_{n+1})=(a_{n+1})(n+1)\]
Then, by the ratio test,
\[\lim_{n \rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|=\lim_{n \rightarrow \infty}\left| \frac{a_{n+1}}{a_{n+1}(n+1)} \right|=\lim_{n \rightarrow \infty}\frac{1}{n+1}=0\] which is smaller than 1, so converges by the ratio test?@Mathematics

Answer 1

Close, but not quite...although yes the series does converge.

Answer 2

\[a_{n}=(a_{n+1})(n+1)\]
So,

\[a_{n+1}=\frac{(a_{n})}{(n+1)}\] so
\[a_{0}=1, \; \; a_{1}=\frac{1}{2}\;\;, \;\; a_{2}=\frac{1}{2*3}, \; \; a_{4}=\frac{1}{2*3*4}\;\; .....\;\;a_{n}=\frac{1}{n!}\]
as n goes to infinity of course a_n converges...and quite fast, to 0.

Answer 3

if you shift the index by 1, I began from "0" SORRY, it is still the same sequence just that
\[a_{1}=1 \;\;\; a_{2}=\frac{1}{2} \;\;\; a_{3}=\frac{1}{2*3} \;\;\;\;\; a_{4}=\frac{1}{2*3*4} \;\;\;\;\; .....a_{n+1}=\frac{1}{(n+1)!} \;\; \]
so it still converges at the same speed to the same number.