# Exact sum of a series

Guidenable

## Homework Statement

Find the exact sum of:

$\sum$ 2/(n7n)

n=1->∞

## The Attempt at a Solution

Let Sn denote the nth partial sum.

ln(Sn) = $\Sigma$ ln(2/(n7n))

=$\Sigma$ ln2 - lnn - nln7

= nln2 - (ln1 + ln7 + ln2 + 2ln7 + ln3 + 3ln7 ...)
= nln2 - ln(n!) - ln7$\Sigma$ n
= ln (2n)/n! - ln7 ($\Sigma$ n)

I'm not sure where to go from here, or if I'm even going in the right direction.

Homework Helper
Gold Member
Do you know how these two power series relate to each other?
$$\sum z^{n-1}$$
$$\sum \frac{1}{n} z^n$$

Guidenable
No, but now I intend to find out. A hint would be appreciated though.

Guidenable
Never mind. Those two series are related by a factor of 1/zn. I should have looked a bit closer :/.

Homework Helper
Gold Member
Hint: $\int z^{n-1} dz = \frac{1}{n} z^n$

Staff Emeritus
Homework Helper

## Homework Statement

Find the exact sum of:

$\sum$ 2/(n7n)

n=1->∞

## The Attempt at a Solution

Let Sn denote the nth partial sum.

ln(Sn) = $\Sigma$ ln(2/(n7n))

=$\Sigma$ ln2 - lnn - nln7

= nln2 - (ln1 + ln7 + ln2 + 2ln7 + ln3 + 3ln7 ...)
= nln2 - ln(n!) - ln7$\Sigma$ n
= ln (2n)/n! - ln7 ($\Sigma$ n)

I'm not sure where to go from here, or if I'm even going in the right direction.
This approach won't work because the very first step is wrong. It's not true that ln(a+b) = ln a + ln b.

Homework Helper
Dearly Missed

## Homework Statement

Find the exact sum of:

$\sum$ 2/(n7n)

n=1->∞

## The Attempt at a Solution

Let Sn denote the nth partial sum.

ln(Sn) = $\Sigma$ ln(2/(n7n))

=$\Sigma$ ln2 - lnn - nln7

= nln2 - (ln1 + ln7 + ln2 + 2ln7 + ln3 + 3ln7 ...)
= nln2 - ln(n!) - ln7$\Sigma$ n
= ln (2n)/n! - ln7 ($\Sigma$ n)

I'm not sure where to go from here, or if I'm even going in the right direction.

Stop right there. What you have done is invalid: the log of a sum is not the sum of the logs. For example, if you claim that log(2 + 3) = log(2) + log(3) you would have log(2) + log(3) = log(2*3) = 6 (because the log of a product is the sum of the logs), so your method would give log(5) = log(6), hence 5 = 6.

RGV

Guidenable
Thanks for the correction, I normally know log rules, just that one slipped. You guys mean bringing the log into the sum, right?

Guidenable
Hint: $\int z^{n-1} dz = \frac{1}{n} z^n$

So I just evaluate the integral?

Homework Helper
Gold Member
So I just evaluate the integral?
Yes, do you see how this can be applied to this problem?

Homework Helper
Dearly Missed
Thanks for the correction, I normally know log rules, just that one slipped. You guys mean bringing the log into the sum, right?

I thought that was what I said.

RGV

Guidenable
Yes, do you see how this can be applied to this problem?

Yes, I understand now, thank you very much!

Eats Dirt
Yes, do you see how this can be applied to this problem?

Hey I was just wondering how you can take the integral of a series to find the sum? Don't series only include integers and will not add up to the same thing as a integral? For example my calculus textbook says when doing the integral test specifically states that the integral is not the sum of the series. Can someone please tell me what I am missing? Thank you.

Homework Helper
Gold Member
Hey I was just wondering how you can take the integral of a series to find the sum? Don't series only include integers and will not add up to the same thing as a integral? For example my calculus textbook says when doing the integral test specifically states that the integral is not the sum of the series. Can someone please tell me what I am missing? Thank you.

If
$$f(z) = \sum a_n z^n$$
then, under certain conditions, you can perform manipulations like this:
$$\int f(z) dz = \int \sum a_n z^n dz = \sum \int a_n z^n dz = \sum \frac{a_n}{n+1} z^{n+1} dz$$
You need to provide justification for interchanging the order of integration and summation. Uniform convergence of the power series is a sufficient condition.

This kind of trick is useful for finding sums such as $\sum (a_n / n) z^n$, provided you have a closed-form expression for $\sum a_n z^n$. Term by term differentiation of the power series can also be useful in some situations.

Staff Emeritus