# Find the set values of x so that the geometric series `1+e^x + e^2x +...` converges.Find the exact value of x so that the sum of infinity of this series is 2.

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### 1 Answer

(a) A geometric series converges if the absolute value of the common ratio is less than 1.

The common ratio of the given series is:

`r=e^x/1=e^(2x)/e^x=e^x`

To solve for values of x that make the series to converge, let |r|<1 .

`|e^x|lt1`

Using the properties of absolute value, we have

`e^xlt1` and `e^xgt-1`

`xltln1` `xgtln (-1)` (Invalid logarithm)

`xlt0`

**Hence, the geometric series `1+e^x+e^(2x)+..` converges when `xlt0` **.

(b) `S_n= 1+ e^x+e^(2x)+...=2`

To solve for x, note that the formula for the sum of infinite geometric series is:

`S_n =sum_(n=0)^oo ar^n = a/(1-r)`

where a is the first term and r is the common ratio.

Substitute `a=1` , `r=e^x` and `S_n=2` .

`2 = 1/(1-e^x)`

`2-2e^x=1`

`2e^x=1`

`e^x=1/2`

`x=ln(1/2)`

`x=-ln2`

**Hence, the value of x in the infinite geometric series `1+e^x+e^(2x)+...=2` is `x=-ln2` .**

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