Question

Plz ans ..... 50
if a_n = n/(n+1)! then find the sum of ∑a_n
n=1

Answer 1

have to find the radius of convergence of the power series ∑anzn∑anzn where an=an= number of divisors of n50n50.

Options available are:

11

5050

150150

00

Please suggest how to proceed.

Using the fact that d(n)≤nd(n)≤n, we have d(n50)≤n50d(n50)≤n50.

Using the First Comparison Test, the series on the right converges if ∣z∣<1∣z∣<1 and however does not converge if ∣z∣≥1∣z∣≥1 and hence so does the series on the left. Radius of Convergence is 1.

[Now for an=nan=n, then R=limn→∞anan+1=1.R=limn→∞anan+1=1.

So,the series on the right converges for ∣z∣<1∣z∣<1]

Now if z=1z=1, then the series ∑anzn∑anzn takes the form ∑d(n50)∑d(n50) and since

d(n50)≥1d(n50)≥1, by evoking the comparison test again, the series diverges for z=1z=1.