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

tinfoQ 2021-06-02 Answered
Find the value of x for which the series converges
\(\sum_{n=1}^\infty(x+2)^n\) Find the sum of the series for those values of x.

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

d2saint0
Answered 2021-06-03 Author has 28286 answers

Consider the series \(\sum_{n=1}^\infty(x+2)^n\)
Let \(a_n=(x+2)^n,\ a_{n+1}=(x+2)^{n+1}\)
\(\frac{a_{n+1}}{a_n}=\frac{(x+2)^{n+1}}{(x+2)^n}\)
\(=x+2\)
\(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}(|x+2|)=|x+2|\)
Using ratio test, this will converge, when
\(|x+2|<1\)
\(-1<(x+2)<1\)
\(-3\)
So, the series is converges at the values lies in the interval, (-3,-1)
And \(\sum_{n=1}^\infty(x+2)^n=(x+2)^1+(x+2)^2+(x+2)^3+...+(x+2)^\infty\)
\(=\frac{x+2}{1-(x+2)}\)
\(=\frac{x+2}{-1-x}\)

Not exactly what you’re looking for?
Ask My Question
46
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...