Question

Determine whether the series converges, and if so find its sum.

(i) \sum^{\infty}_{k=1}\frac{1}{9k^2+3k-2}

(ii) \sum^{\infty}_{k=1}\frac{4^{k+2}}{7^{k-1}}

(iii) \sum^{\infty}_{k=1}(\frac{1}{(k+p)(k+p+1)})

Answer 1

can you post the real quest... i did not get it

Answer 2

yup i understand the series look

Answer 3

have u heard about telescopic series

Answer 4

if so, you can apply it to find out it converges to a number

Answer 5

(i) $$\sum^{\infty}_{k=1}\frac{1}{9k^2+3k-2}$$

(ii)$$ \sum^{\infty}_{k=1}\frac{4^{k+2}}{7^{k-1}}$$

(iii)$$ \sum^{\infty}_{k=1}(\frac{1}{(k+p)(k+p+1)})$$

Answer 6

ops too many... am talking about the first

Answer 7

thank you "Perl" . it seems that "Krishtika" is not listening :)

Answer 8

hint:
we can rewrite the first sum as below:
\[\Large \frac{1}{3}\sum\limits_{k = 1}^{ + \infty } {\left( {\frac{1}{{3k - 1}} - \frac{1}{{3k + 2}}} \right)} \]

Answer 9

i'm back
so u people got the solution

Answer 10

\[
\sum^{\infty}_{k=1}\frac{4^{k+2}}{7^{k-1}}=16\times7\sum^{\infty}_{k=1}\frac{4^{k}}{7^{k}}
\]
The third one looks convergent to me but I have no idea how to evaluate the sum.

Answer 11

Partial fraction decomposition on the third one just like the first one. This will yield a telescoping series.

Answer 12

for 3) make comparison test with \(\dfrac{1}{k^2}\), we can get the sum is convergent.
to find the sum, use limit
\[lim_{t\rightarrow \infty}\int_1^t \dfrac{1}{k+p}-\dfrac{1}{k+p+1}dk\]
After calculating, I got 1 :)

Answer 13

that method is an advanced one compared to Telescopic series partial fraction way....the second is much easier

Answer 14

@Loser66, comparing the sum to the integral doesn't tell us much about the actual sum, but rather an upper (or lower, depending on the summand's behavior) bound.