#converging infinite sum
9 messages · Page 1 of 1 (latest)
Prove that $\sum_{k=1}^n \frac{k^3}{2^k}=26-\frac{n^3+6n^2+18n+26}{2^n}$, then just take the limit
you mean k^3/2^k instead right?
and also how did you get that
just curious
@untold storm
Yes I did
SWR