1. ## Sum of series .. keep getting wrong answer

Hello,
Question :
Find the sum of the convergent series ( (1/e)^n + ( 1 / n(n+1) ) ) from n=1 to n=infinity
Clearly, the series is a sum of two convergent series
First series is geometric series with sum = a / (1-r) = 1 / ( 1 - (1/e) ) = e / (e-1)
Second series is telescoping series with Sn = 1 - ( 1 / (n+1) ) which approach 1 as n approach infinity
So the sum is the sum of the two series 1 + ( e/(e-1) )
But the book and wolframalpha keep saying that the sum is e/(e-1) !

2. ## Re: Sum of series .. keep getting wrong answer

Originally Posted by Ted
Hello,
Question :
Find the sum of the convergent series ( (1/e)^n + ( 1 / n(n+1) ) ) from n=1 to n=infinity
Clearly, the series is a sum of two convergent series
First series is geometric series with sum = a / (1-r) = 1 / ( 1 - (1/e) ) = e / (e-1)
Second series is telescoping series with Sn = 1 - ( 1 / (n+1) ) which approach 1 as n approach infinity
So the sum is the sum of the two series 1 + ( e/(e-1) )
But the book and wolframalpha keep saying that the sum is e/(e-1) !
Because the telescoping sequence isn't defined for n=0, you are summing from 1 to infinity. Not 0 to infinity.

$\displaystyle{\sum_{n=1}^{\infty}}\left(\dfrac 1 e\right)^n = \dfrac 1 {e-1}$

$1 + \dfrac 1 {e-1} = \dfrac e {e-1}$

3. ## Re: Sum of series .. keep getting wrong answer

Originally Posted by romsek
Because the telescoping sequence isn't defined for n=0, you are summing from 1 to infinity. Not 0 to infinity.
$\displaystyle{\sum_{n=1}^{\infty}}\left(\dfrac 1 e\right)^n = \dfrac 1 {e-1}$
I know that why does it start from n=1
How did you calculate that sum ?
Applying geometric series formula gives e / (e-1)

4. ## Re: Sum of series .. keep getting wrong answer

Originally Posted by Ted
I know that why does it start from n=1
How did you calculate that sum ?
Applying geometric series formula gives e / (e-1)
it gets $\dfrac e {e-1}$ when you start from $n=0$

it gets $\dfrac 1 {e-1}$ when you start from $n=1$

5. ## Re: Sum of series .. keep getting wrong answer

Why is that?
My professor did not mention that
just give us the form of geometric series a * (r)^n and the formula of the sum a / (1-r)

6. ## Re: Sum of series .. keep getting wrong answer

Originally Posted by Ted
Why is that?
My professor did not mention that
just give us the form of geometric series a * (r)^n and the formula of the sum a / (1-r)
you can wiki all this faster than I can type it out and have you argue about it with me.

7. ## Re: Sum of series .. keep getting wrong answer

$\displaystyle \sum_{n=1}^\infty r^n = \left(\sum_{n=0}^\infty r^n\right) - r^0 = \left(\sum_{n=0}^\infty r^n\right)-1 = \dfrac{1}{1-r}-1$

In your case, $\displaystyle r = \dfrac{1}{e}$, so you have:

$\displaystyle \sum_{n=1}^\infty \left(\dfrac{1}{e}\right)^n = \dfrac{1}{1-\tfrac{1}{e}} - 1 = \dfrac{e}{e-1} - 1 = \dfrac{1}{e-1}$

Just as romsek said.

8. ## Re: Sum of series .. keep getting wrong answer

All what do you need to say is that the formula apply when n starts with 0 and i will get it without any arguing.
Thank you all