# College Calculus

posted by .

Suppose the series An (from n=1 to INF) is known to be convergent. Prove that series 1/(An) (from n=1 to INF) is a divergent series.

• College Calculus -

If the series is convergent then you know that An tends to zero, therefore 1/An tends to infinity and thus the second series cannot converge

## Similar Questions

2. ### calculus

is this correct? use the integral test to determine if this series is convergent or divergent: the series from n=2 to infinity of 1/(n*square root of (ln(n))) I said it was divergent because the integral went to infinity
3. ### Pre-calculus

Which of the following series is divergent?
4. ### Calculus

series: n=1 to n=inf of 6/(n^3-4) convergent or divergent?
5. ### calculus

With power series, is an endpoint convergent if you plug it back into the original series, and get an alternating series that is conditionally convergent?
6. ### Calculus

Show that the following series is absolutely convergent: Summation from 1 to infinity: [(-1)^n * (n+1) * 3^n]/ [2^(2n+1)] I've done the ratio test and replaced n in this series with n+1. I get 3/4 in the end, which is less than 1, …
7. ### Calculus 2

Hello, I don't know what test to use for this series: Determine the sum of the following series: inf E n=1 (2^n + 9^n) / 12^n thank you!
8. ### Calculus 2 (Series - Convergent or Divergent?)

Can someone show me a step by step process and explanation how to solve this problem?
9. ### Integral Calculus - Series

Find if series is convergent or divergent. Series from n=2 to infinity (4n+7)/(3n^3 -8n)
10. ### Calculus II

I need to find if the summation of (n^4)/(n^10 + 1) is convergent or divergent from n=1 to infinity. I tried splitting it up into two sums, one being 1/n^6, which would be convergent because p=6>1, and then the other being n^4, …

More Similar Questions