1 Answer | Add Yours
(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:
To solve for values of x that make the series to converge, let |r|<1 .
Using the properties of absolute value, we have
`e^xlt1` and `e^xgt-1`
`xltln1` `xgtln (-1)` (Invalid logarithm)
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)`
Hence, the value of x in the infinite geometric series `1+e^x+e^(2x)+...=2` is `x=-ln2` .
We’ve answered 319,811 questions. We can answer yours, too.Ask a question