- #1
dobry_den
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Hi, could you please check if my solution is correct?
Test the following series for convergence:
\sum_{n=1}^{\infty}\frac{1!+2!+...+n!}{(\left 2n \right)!}
I can use a slightly altered series
\sum_{n=1}^{\infty}\frac{nn!}{(\left 2n \right)!}
whose every term is >= than the corresponding term in the original series.. and thus if this altered series converges, then the original one should so as well...
Then, if I use the limit ratio test for the second series:
\lim_{n \rightarrow \infty}\frac{(\left n+1\right)(\left n+1 \right)!}{(\left 2n+2 \right)!}\frac{(\left 2n\right)!}{n(\left n \right)!} = 0
This means that the altered series is convergent, and thus the original series is also convergent.
Is this reasoning correct? Thanks in advance!
Homework Statement
Test the following series for convergence:
\sum_{n=1}^{\infty}\frac{1!+2!+...+n!}{(\left 2n \right)!}
The Attempt at a Solution
I can use a slightly altered series
\sum_{n=1}^{\infty}\frac{nn!}{(\left 2n \right)!}
whose every term is >= than the corresponding term in the original series.. and thus if this altered series converges, then the original one should so as well...
Then, if I use the limit ratio test for the second series:
\lim_{n \rightarrow \infty}\frac{(\left n+1\right)(\left n+1 \right)!}{(\left 2n+2 \right)!}\frac{(\left 2n\right)!}{n(\left n \right)!} = 0
This means that the altered series is convergent, and thus the original series is also convergent.
Is this reasoning correct? Thanks in advance!
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