Question

\[\sum_{k=1}{\infty} \frac{k+5}{5^{k}}\]

Answer 1

\sum_{k=1}^{\infty} \frac{k+5}{5^{k}}

Answer 2

\[\sum_{k=1}^{\infty} \frac{k+5}{5^{k}}\]

Answer 3

convergence or divergence problem

Answer 4

I imagine the ratio test would work well.

Answer 5

ok i'll try it

Answer 6

\[ \frac{k+1+5}{5^{k+1}}*\frac{5^{k}}{k+5}\]
\[ \frac{5^{k}}{5^{k+1}}*\frac{k+6}{k+5}\]
\[ 5*\frac{k+6}{k+5}\]

Answer 7

Shouldn't that be\[\frac{k+6}{5(k+5)}\]

Answer 8

i don't think so, but i could be wrong

Answer 9

k-k-1 is -1 so yes you are right

Answer 10

so since the denominator is greater is it divergent?

Answer 11

Careful. To use the ratio test, we need to look at \[\lim_{k\rightarrow\infty} \frac{k+6}{5k+25}=L\]If \(L<1\) it's convergent, if \(L>1\) it diverges, if \(L=1\), it tells us nothing.

What is \[\lim_{k\rightarrow\infty} \frac{k+6}{5k+25}?\]

Answer 12

if we multiply the numerator and dinominator by 1/k and let k-> infty we get 6/25

Answer 13

I believe it would be 1/5 and not 6/25, but in either case, this is less than 1, so the series converges.

Answer 14

i think you're right....but why 1/5 sir George

Answer 15

\[\lim_{k\rightarrow\infty} \frac{k+6}{5k+25}\cdot \frac{1/k}{1/k}=\lim_{k\rightarrow\infty} \frac{1+6/k}{5+25/k}\]Since \(6/k\) and \(25/k\) go to 0, we get 1/5.

Answer 16

i'm soo silly thanks

Answer 17

you're welcome.

Answer 18

it's late where i'm from...i'm forgetting basic algebra

Answer 19

I've done much worse. For example, I could have sworn 9+4=11 not too long ago.

Answer 20

lolz it happens to the best of us :)