$\displaystyle \frac{1}{n}-\frac{-1^n}{n^2}<0.000001$

can i solve this algebraically?

if not, how do i solve it?

many thanks. (Nod)

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- Nov 28th 2011, 12:17 PMBabyMiloLog problem.
$\displaystyle \frac{1}{n}-\frac{-1^n}{n^2}<0.000001$

can i solve this algebraically?

if not, how do i solve it?

many thanks. (Nod) - Nov 28th 2011, 01:45 PMAmerRe: Log problem.
$\displaystyle \frac{n - (-1)^n}{n^2} < \frac{1}{1000000} $

take n even 2k instead of n

$\displaystyle \frac{2k - 1}{4k^2} < \frac{1}{1000000}$

can you solve it ??

try n odd sub 2k+1 instead of n - Nov 29th 2011, 10:23 AMBabyMiloRe: Log problem.
for even, k>499999

for odd, k>500000

how to i proceed?

how answer should be 1000002.

i can see if i add the two ks, it comes to k>1000001

is this the way to do it?

many thanks. :) - Nov 29th 2011, 10:29 AMQuackyRe: Log problem.
for even, $\displaystyle k>499999$ So $\displaystyle \frac{n}{2}>499999$ remembering that $\displaystyle n=2k$

for odd, $\displaystyle k>500000$ So $\displaystyle \frac{n-1}{2}>500000$ remembering that $\displaystyle n=2k+1$

What is the solution which satisfies for all $\displaystyle n$?