Question

In this exercise, determine the convergence or divergence of the series.

n=1(1)nnn+1\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt{n}}{n+1}

Solution

Verified
Answered 1 year ago
Answered 1 year ago
Step 1
1 of 2

The given series:

n=1(1)nnn+1\sum_{n=1}^\infin\frac{(-1)^n\sqrt n}{n+1}

To apply the alternating series test, first lets define bnb_n:

n=1(1)nnn+1=n=1(1)nnn+1bn=nn+1\sum_{n=1}^\infin\frac{(-1)^n\sqrt n}{n+1}=\sum_{n=1}^\infin(-1)^n\frac{\sqrt n}{n+1}\longmapsto b_n=\frac{\sqrt n}{n+1}

Create an account to view solutions

Create an account to view solutions

More related questions

1/4

1/7