# Compute each sum below. Give exact values, not decimal approximations. If the sum does not exist, click on "No sum".?

### 5 Answers

- geezerLv 71 month ago
Why do people asking other people to do their homework for them NEVER say ''Please'' ?

Thumb down !

I've read your question again and NO .. I can't see the word PLEASE there !!

- Geeganage WLv 51 month ago
Hope that you are expecting the sum to infinity in the series, (4/9)+(4/9)^2+...,

Let give a little modification by adding 1 and substracting the same.

Then, -1 + 1+ (4/9)+(4/9)^2+...,

1+ (4/9)+(4/9)^2+...,is a geometric series & the common ration, r = 4/9 <1, and a = 1

Therefore sum to infinity = a/(1-r) = 1/[1-4/9] = 9/5.

Thus the sum of the given series = 9/5 - 1= 4/5.

Second series is also a geometric series if you add 1. Then r = -4 <1 and a = 1,

As above sum to infinity = a/(1-r) = 1/(1+4) = 1/5

Therefore sum to infinity of the given series = 1/5 - 1 = - 4/5.

- Jim MoorLv 71 month ago
Sum (4/9)^n = 4/5

But, you should learn the technique, not ask us to do your homework

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- az_lenderLv 71 month ago
The first one is (4/9)/(1 - 4/9) = 4/5.

The second one diverges, i.e., "No sum"