Find the sum of the following convergent series: a)∑_(n=1)^∞ n/(n+1)! b) (∑_(n=1)^∞ 2/5^n)-(∑_(n=1)^∞ 3^n/5^n)

Question

Can you help me with this problem step by step. Usually I dont have problems with these but now I don't understand nothing

Answer 1

For part b)

The formula for the sum of an infinite geometric series is a/(1-r) where a is the first term of the series and r is the common ratio which is less than 1 in absolute value.

Think of the first sum as 2 (1/5 + 1/25 + 1/125+...). The series in parentheses is a geometric series with a = 1/5 and r = 1/5. so its sum is (1/5 / (1 - 1/5) = 1/4. Multiply that by 2 to find that the first sum is 1/2.

The 2nd sum is (3/5 + 9/25 + 27/125+...), a geometric series with a = 3/5 and r = 3/5 so its sum is:

3/5 / (1 - 3/5) = 3/2. So the expression in part b evaluates to 1/2 - 3/2 = -1.

Check here for confirmation:

desmos.com/calculator/xrkxovbt8m

Answer 2

To see the convergence of the given series you can use some tests. For example, for the first series you can use the ratio test

If r < 1, then the series is convergent. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

In your example

a_n=n/(n+1)! , then a_n+1=(n+1)/(n+2)!  and

a_n+1/a_n=(n+1)/(n+2)!*((n+1)!)/n= (n+1)/(n(n+2))-> 0, when n-> ∞. So the first series converges.

Similar for the b)