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from n=1 {sum} to infinity of 1 / [2 + (5)^n]

2007-07-15 17:09:03 · 5 answers · asked by beeboppinbippity 1 in Science & Mathematics Mathematics

5 answers

This series converges. The series is monotonic and each term is less than that in the series

n=1 {sum} to infinity of 5^(-n),
= (1/5) * k = 0 {sum} to infinity 5^(-k) which converges to (1/5) * [1/(1 - 1/5)] = 1/4

The series in the problem therefore is bounded on the bottom by zero and on the top by 1/4. The bound on the bottom gets raised as you start evaluating the individual terms in the series. The top bound lowers by the difference between the series in question and the sum of the same number of terms of the series 1 / 5^n.

2007-07-15 17:32:06 · answer #1 · answered by devilsadvocate1728 6 · 0 0

1/[2+5^n) < 1/5^n and 1/5^n is the general term of a geometric series with r=1/5 < 1 (convergent) ( nothing further needed)

2007-07-16 09:03:10 · answer #2 · answered by santmann2002 7 · 0 0

this series is converges because
if limit as n approches infinity the series becomes as

1/[2+infinity]
=1/ (infinity)=0
i-e
series is converges

2007-07-16 01:48:49 · answer #3 · answered by usman 1 · 0 0

First answer is not correct; it applies to a sequence, not a series. I beliveve that this series converges, but I don't see a way to prove it, or to determine what the limit might be.

2007-07-16 00:24:05 · answer #4 · answered by Anonymous · 0 0

converge to zero!

any number n will make the denominator much larger than the numerator. and if this n approaches infinity, the resulted function will be equal to zero.

2007-07-16 00:13:34 · answer #5 · answered by Mr. Engr. 3 · 0 0

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