Question

what will be the closed formula of the series S := 14/(49k^2 +56-33) from k=0 to infinity? Thanks!

Answer 1

is the denominator 56-33 or is it 56x - 33

Answer 2

im sorry it should be 56k

Answer 3

yeah figured. partial fractions yes?

Answer 4

that is why the 14 is there to make it nice

Answer 5

yep i ended up with 1/7k-3 and 1/7k+11

Answer 6

they are being subtracted

Answer 7

o.k i'll outline a method called the tn method dived ,using partial fractions, the term into 2 such the [t_n] ad [t_n+1] terms either partially or wholly can cell out.

Answer 8

get
\[\sum_{k=1}^{\infty}\frac{1}{7k-3}-\frac{1}{7k+11}\]

Answer 9

should it be from k=0 or k=1?

Answer 10

By partial fractions,\[S = \sum_{k = 0}^\infty \left(\frac{1}{7k-3}-\frac{1}{7k+11}\right)\]which is a telescoping sum and the result is\[\frac{1}{7\cdot 0 - 3}+\frac{1}{7\cdot1-3}=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12}.\]

Answer 11

but what will be the closed formula? from the answer above, that means that the series converges on -1/12 right? Thanks!

Answer 12

The numerical result (I assume that's what you mean when you say closed formula) of the series is\[-\frac{1}{12}\]as I said above.

Answer 13

oh, i thought a closed formula will be similar to n(n+1)/2 and such.

Answer 14

That would be true if you had something like\[S_n \doteq \sum_{k = 0}^n \frac{14}{49k^2+56-33}.\]If you have a sum from 0 to infinite there is no 'n', so obviously the result can't be in terms of n.
For the partial sum above, you can do the same as before:\begin{eqnarray*}S_n &=& \sum_{k = 0}^n\left(\frac{1}{7k-3}-\frac{1}{7k+11}\right)\\ &=& \sum_{k = 0}^n\left(\left(\frac{1}{7k-3}-\frac{1}{7(k+1)-3}\right)+\left(\frac{1}{7(k+1)-3} - \frac{1}{7(k+2)-3}\right)\right)\\&=&\frac{1}{7\cdot0-3}-\frac{1}{7(n+1)-3}+\frac{1}{7\cdot1-3}-\frac{1}{7(n+2)-3}\\&=&-\frac{7}{12}\frac{32+39n+7n^2}{(4+7n)(11+7n)}.\end{eqnarray*}

Answer 15

what happened to 1/7k+11?

Answer 16

I wrote\[\frac{1}{7k+11}\]as\[\frac{1}{7(k+2)-3}\]to make the telescoping sum clearer.

Answer 17

oh i see thanks!