Integrating for approximation of a sum

[goraemon]

Homework Statement

Find an N so that $∑^{\infty}_{n=1}\frac{log(n)}{n^2}$ is between $∑^{N}_{n=1}\frac{log (n)}{n^2}$ and $∑^{N}_{n=1}\frac{log(n)}{n^2}+0.005.$

Homework Equations

Definite integration

The Attempt at a Solution

I began by taking a definite integral: $\int^{\infty}_{N}\frac{log(n)}{n^2}dn$ and, using integration by parts, arrived at the following answer: $\frac{log(N)+1}{N}$. (Is this right? If not, I could post the steps I used to try to see where I made an error)

Next we need $\frac{log(N)+1}{N}$ to be within 0.005 as given by the problem, so:

$\frac{log(N)+1}{N}=0.005=\frac{1}{200}$

But I'm having trouble how to solve for N algebraically. Would appreciate any help.

 

[BiGyElLoWhAt]

I actually got -1 times what you have. remember: $\int \frac{1}{n^2} = -\frac{1}{n}$

 

[goraemon]

I actually got -1 times what you have. remember: $\int \frac{1}{n^2} = -\frac{1}{n}$

Right but since we're taking a definite integral on the interval from N to ∞, and since $\frac{-log(N)-1}{N}$ as N approaches infinity equals zero, shouldn't the definite integral work out to:
$0-\frac{-log(N)-1}{N}=\frac{log(N)+1}{N}$?

 

[pasmith]

Homework Statement

Find an N so that $∑^{\infty}_{n=1}\frac{log(n)}{n^2}$ is between $∑^{N}_{n=1}\frac{log (n)}{n^2}$ and $∑^{N}_{n=1}\frac{log(n)}{n^2}+0.005.$

Homework Equations

Definite integration

The Attempt at a Solution

I began by taking a definite integral: $\int^{\infty}_{N}\frac{log(n)}{n^2}dn$ and, using integration by parts, arrived at the following answer: $\frac{log(N)+1}{N}$. (Is this right? If not, I could post the steps I used to try to see where I made an error)

Your integral is correct.

Next we need $\frac{log(N)+1}{N}$ to be within 0.005 as given by the problem, so:

$\frac{log(N)+1}{N}=0.005=\frac{1}{200}$

The graph of x^{-2}\log(x) has a maximum in [1,2], so it's difficult to say whether the integral is an over-estimate or under-estimate of the sum. To be safe I'd require that \frac{\log(N) + 1}{N} < \frac{1}{400}.

But I'm having trouble how to solve for N algebraically. Would appreciate any help.

Equations of that type have to be solved numerically. You're looking for zeroes in x &gt; 0 of f(x) = \frac{x}{200} - 1 - \log x. The derivative of this function is <br /> f&#039;(x) = \frac{1}{200} - \frac{1}{x} so the second derivative (x^{-2}) is everywhere positive. There is a minimum at x = 200 where f(200) &lt; 0. Since f(x) \to +\infty as x \to 0 or x \to \infty there are exactly two solutions. Logic dictates you want the larger, since if N works then any larger M should also work.

 

[goraemon]

Thanks, I tried solving it numerically and came up with N = 1687, which the textbook confirms is correct.