# Sum of geometric series

For the question, shouldn't the sum be a(1/1-r) since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?

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LCKurtz
Homework Helper
Gold Member
Isn't the first term, which you are calling $a$, equal to $r$ in that sum?

Isn't the first term, which you are calling $a$, equal to $r$ in that sum?
Isn't $a = \frac {1}{4}$ and $r = \frac{2}{e}$?

Also, for this question, should the interval of convergence be $[3,3]$? Why exactly did they say $(-1,1)$?

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Mark44
Mentor
Isn't $a = \frac {1}{4}$ and $r = \frac{2}{e}$?
No. a = 1/(2e)

Last edited:
Mark44
Mentor
For the question, shouldn't the sum be a(1/1-r)
When you write fractions as text on a single line, you need to be careful about parentheses. 1/1-r means (1/1) - r, and not 1/(1 - r) as you intended.

You have it below, as well.
since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?

LCKurtz
Homework Helper
Gold Member
Isn't $a = \frac {1}{4}$ and $r = \frac{2}{e}$?
In the picture where he has written $\frac 1 4\sum_{k=1}^\infty r^k$ the 1/4 is factored out and the remaining sum has $a=r=\frac 2 e$.

Ray Vickson
Homework Helper
Dearly Missed
For the question, shouldn't the sum be a(1/1-r) since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?
Of course there is a difference! If you start the sum at $n = 0$ the first term is $a$; if you start it at $n = 1$ the first term is $ar$. The two sums differ by $a$.

Anyway, they didn't write $a(r/1-r) = ar - ar = 0$; they wrote $ar/(1-r)$, where parentheses make all the difference in the world.

Fredrik
Staff Emeritus
Gold Member
The sum ($\sum_k r^k$) is 1/(1-r) when the summation starts at 0. If it starts at 1, you can do this: $$\sum_{k=1}^\infty r^k =r\sum_{k=1}^\infty r^{k-1} =r\sum_{k=0}^\infty r^k.$$

The sum ($\sum_k r^k$) is 1/(1-r) when the summation starts at 0. If it starts at 1, you can do this: $$\sum_{k=1}^\infty r^k =r\sum_{k=1}^\infty r^{k-1} =r\sum_{k=0}^\infty r^k.$$
So if it is instead $∑_{n=3}^∞r^n$ then this equals $∑_{n=0}^∞r^{n+3} = r^3∑_{n=0}^∞r^{n}$ so then the sum is $\frac{r^3}{1-r}$ correct?

LCKurtz
So if it is instead $∑_{n=3}^∞r^n$ then this equals $∑_{n=0}^∞r^{n+3} = r^3∑_{n=0}^∞r^{n}$ so then the sum is $\frac{r^3}{1-r}$ correct?
Yes. This agrees with the formula $\frac a {1-r}$ since the first term, $a$, in $\sum_{n=0}^∞r^{n+3}$ is $r^3$, as is the first term in $\sum_{n=3}^∞r^n$. It doesn't matter how you write it.