Question

1/(2n)^2 this converges to? or not converge?

Answer 1

The elements get small fast, so it likely converges. Handbooks would have series expressions, one of which might match this. I do not recall how to derive the answer, however.

Answer 2

Compare to a p-series that, which converges for n > 1: http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#P-series

Since for p-series, n=2 this function is 1/(2n)^2 < 1/n^2, which converges, so does 1/(2n)^2 by the comparison test: http://en.wikipedia.org/wiki/Direct_comparison_test

Answer 3

i have 4 options it is converges to 0 ot 1/2 or 1/4 or not converge?

Answer 4

Since it converges, what is the limit as \(n\to\infty\) of \(\cfrac{1}{(2n)^2}\)

Answer 5

what is \(\cfrac{1}{\infty}\)?

Answer 6

yaa n->infinite

Answer 7

$$
\lim_{n\to\infty}\cfrac{1}{(2n)^2}\to\cfrac{1}{\infty}=0
$$

Answer 8

1/infinity is infinitly small so it can be treated as zero

Answer 9

yes, it approaches the asymptote y=0.

Answer 10

mm yaa
your answer it is converges to zero?