series convergence tests: sum of (-1)^n/(3^n*n!)

[paulxzt]

1) Approximate the sum of the series correct to 4 decimal places

Sum of (-1)^n / (3^n*n!) with n = 1 to infinity

2) Determine whether the series is absolutely convergent

Sum of (-1)^n * n / (n^2+1) with n = 1 to infinity
do i take the limit of (n/n^2+1) ?

any help with these? I am absolutely lost, thanks

 

[skeeter]

Re: series convergence tests

Approximate the sum of the series correct to 4 decimal places

Sum of (-1)^n / (3^n*n!) with n = 1 to infinity

Determine whether the series is absolutely convergent

the series converges absolutely since, term for term, 1/(3ⁿ*n!) < 1/3ⁿ, a known convergent geometric series.

since the series alternates, the total error < the first omitted term in the sum. so ...
you want to find the value of n such that 1/(3ⁿ*n!) < 1/10000, then find the partial sum from 1 to n-1 to get the 4 decimal place accuracy estimate.