Question

find the value of x for which the series converges

sigma (x+7)^n for n=1 to n=infinity

Answer #1

According to ratio test, let be a given series. The series converges if p<1 where

Here the series is

For the series to be convergent, p<1.

So, **for the value of x from -8 to -6, the series
converges**.

Given: Sigma (infinity) (n=1) sin((2n-1)pi/2)ne^-n
Question: Determine if the series converges or diverges
Additional: If converges, is it conditional or absolute?

find the radius of convergence, R, of the series.
Sigma n=1 to infinity x^n/(4^nn^5)
R=
Find the interval, I of convergence of the series.

find the sum of the series
sigma n=0 to infinity {2 [3^(n/2 -2) / 7^(n+1)] + sin ( (n+1)pi
/ 2n+1) - sin (n pi / 2n-1)}

Sigma(1, infinity) (2^n x^n)/(n!) : Radius of convergence of
this series?

find the sun of the series
sigma n=0 to infinity 2(3^n/2 -2)/7^n+1 + sin ( (n+1)pi / 2n+1)
- sin (npi/2n-1)

Infinity Sigma n=1 (n+1 / n^7/3 + sqrt n)
Does this series converge or diverge?

How do I show whether sigma(n=1 to infinity) sqrt(n)/(1+n^2)
converges or not?

suppose sigma n=1 to infinity of square root ((a_n)^2 +
(b_n)^2)) converges. Show that both sigma a_n and sigma b_n
converge absolutely.

Determine if the series converges conditionally, converges
absolutely, or diverges.
/sum(n=1 to infinity) ((-1)^n(2n^2))/(n^2+4)
/sum(n=1 to infinity) sin(4n)/4^n

Show that the series sum(an) from n=1 to infinity
where each an >= 0 converges if and only if for every
epsilon>0 there is an integer N such that | sum(ak )
from k=N to infinity | < epsilon

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