Approximating an infinite series

**MathIsOhSoHard** · October 8th 2012, 06:41 PM

Approximate a sum for the infinite series:
$\sum_{n=1}^{\infty}\frac{1}{n^4}$
with a maximum error of $0.02$

Answer:
$\sum_{n=1}^{\infty}\frac{1}{n^4}\approx 1.0748$

My attempt:
For every $N\in\mathbb{N}$ :
$\int\limits_{N+1}^t\frac{1}{x^4}\, dx$ $=\left [ \frac{1}{3x^3} \right ]_{N+1}^t=\frac{1}{3(N+1)^3}-\frac{1}{3t^3}$

For $t\rightarrow \infty$ :
$\int\limits_{N+1}^t\frac{1}{x^4}\, dx$ $=\frac{1}{3(N+1)^3}$

$f(x)=\frac{1}{x^4}$
$\int\limits_{N+1}^{\infty}f(x)\, dx$ $=\int\limits_{N+1}^{\infty}\frac{1}{x^4}\, dx$ $=\frac{1}{3(N+1)^3}$
$f(N+1)=\frac{1}{(N+1)^4}$

$\int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3(N+1)^3}+\frac{1}{(N+1)^4}=\frac{ 1}{3}\frac{N+4}{(N+1)^4}$

Expanding we get:
$\int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3}\frac{N+4}{n^4+4N^3+6N^2+4N+1}$ $=\frac{N+4}{N(3N^3+12N^2+18N+12)+3}$

Since:
$N(3N^3+12N^2+18N+12)+3\ge N(3N^3+12N^2+18N+12)$

We get that:
$\int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)$ $\le \frac{N+4}{N(3N^3+12N^2+18N+12)}\le \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}$

Which shows us that:
$\frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}\le 0.02$

However when I try to solve for $N$ I get a negative $N$ -value which doesn't make sense. Can somebody help me get the correct $N$

**TheEmptySet** · October 8th 2012, 07:21 PM

Originally Posted by MathIsOhSoHard

Approximate a sum for the infinite series:
$\sum_{n=1}^{\infty}\frac{1}{n^4}$
with a maximum error of $0.02$

Answer:
$\sum_{n=1}^{\infty}\frac{1}{n^4}\approx 1.0748$

My attempt:
For every $N\in\mathbb{N}$ :
$\int\limits_{N+1}^t\frac{1}{x^4}\, dx$ $=\left [ \frac{1}{3x^3} \right ]_{N+1}^t=\frac{1}{3(N+1)^3}-\frac{1}{3t^3}$

For $t\rightarrow \infty$ :
$\int\limits_{N+1}^t\frac{1}{x^4}\, dx$ $=\frac{1}{3(N+1)^3}$

$f(x)=\frac{1}{x^4}$
$\int\limits_{N+1}^{\infty}f(x)\, dx$ $=\int\limits_{N+1}^{\infty}\frac{1}{x^4}\, dx$ $=\frac{1}{3(N+1)^3}$
$f(N+1)=\frac{1}{(N+1)^4}$

$\int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3(N+1)^3}+\frac{1}{(N+1)^4}=\frac{ 1}{3}\frac{N+4}{(N+1)^4}$

Expanding we get:
$\int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3}\frac{N+4}{n^4+4N^3+6N^2+4N+1}$ $=\frac{N+4}{N(3N^3+12N^2+18N+12)+3}$

Since:
$N(3N^3+12N^2+18N+12)+3\ge N(3N^3+12N^2+18N+12)$

We get that:
$\int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)$ $\le \frac{N+4}{N(3N^3+12N^2+18N+12)}\le \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}$

Which shows us that:
$\frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}\le 0.02$

However when I try to solve for $N$ I get a negative $N$ -value which doesn't make sense. Can somebody help me get the correct $N$

Solving that 4th order equation for N is a nightmare. You can get an approximate value

$3N^4< N(3N^3+12N^2+18+\frac{1}{3}) \iff \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})} < \frac{1}{3N^4}$

This much better to solve

$\frac{1}{3N^4} < \frac{2}{100} \iff 100 < 6n^4$

You can see that $n=3$ works

**MaxJasper** · October 8th 2012, 07:21 PM

$\sum _{k=1}^{\infty } \frac{1}{k^4}$ = $1+\frac{1}{2^4}+\frac{1}{3^4}$

$\frac{1}{3^4}<.02$

**chiro** · October 8th 2012, 08:16 PM

Hey MathIsOhSoHard.

Can you show us how you solved the roots of the equation N(3N^3 + 12N^2) + 18 + 1/3)*0.02 - 1 = 0? (Knowing the roots means you know which values of N will be on either side of the inequality).

Thread: Approximating an infinite series

LinkBack

Thread Tools

Display

Approximating an infinite series

Re: Approximating an infinite series

Re: Approximating an infinite series

Re: Approximating an infinite series

Similar Math Help Forum Discussions

Approximating sin(1) Using Taylor Series

Euler-Maclaurin Series formula on an infinite series?

Approximating definite integral using series

approximating definite integral using power series

calculus II areas about axis, pressure, infinite series, MacLaurin series

Search Tags