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Numerade Educator



Problem 58 Medium Difficulty

Find the values of $ x $ for which the series converges. Find the sum of the series for those values of $ x. $
$ \displaystyle \sum_{n = 1}^{\infty} (x + 2)^n $


Interval of convergence is $x \in(-3,-1)$
The sum of the series is $-\frac{x+2}{x+1}$

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Video Transcript

Were given a series and We have to find the values of X.. this series converges and to find the some of the series For each value of X. So the series is the sum from n equals 1 to infinity of X-plus 2 To the ends. Power Notice that if X is some number, This is the form of a geometric series in particular with initial term X plus two. We'll call this a and common ratio our which is X plus to Also. Now we know that a geometric series converges If and only if The absolute value of the common ratio is less than 1. Therefore this is true if and only if the absolute value of X plus two is less than one, We can solve this inequality for X. This means X plus two must lie between one and negative one. In other words Xmas lie between on the one hand meg three and on the other hand negative one. These are the values of X for which it converges interval notation. X must lie in left parentheses, negative three, negative one right parenthesis. Now, for each of these values of X, we want to find this some. Well, we simply use the formula for the sum of a geometric series. This is the initial term which was X plus two over one minus the common ratio, Which is also X-plus 2. So this gives us X plus two over negative x minus one, Which we can factor out a negative and write this as negative times X plus two over X plus one. And this is the sum of these series.