I have to solve - \sum_{i=1}^\infty(\frac{5}{12})^i - geometric series?

Talamancoeb 2022-01-07 Answered
I have to solve -
\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}\)
- geometric series?

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

rodclassique4r
Answered 2022-01-08 Author has 2682 answers
There is another, faster, explanation, directly linked to the nature of the series: you can factor out x from each term of the series:
\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={x}{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{{i}-{1}}}\)
\(\displaystyle={x}{\sum_{{{i}={0}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}\)
(setting \(\displaystyle{i}\to{i}-{1}\)), so
\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={\frac{{{x}}}{{{1}-{x}}}}\)
Not exactly what you’re looking for?
Ask My Question
0
 
Shannon Hodgkinson
Answered 2022-01-09 Author has 2714 answers
HINT:
\(\displaystyle{\sum_{{{i}={0}}}^{\infty}}{x}_{{i}}={\frac{{{1}}}{{{1}-{x}}}}={x}_{{0}}+{\sum_{{{i}={1}}}^{\infty}}{x}_{{i}}\)
0
karton
Answered 2022-01-11 Author has 8659 answers

Observe that for suitable x:
\((1-x)\sum_{i=k}^\infty x^i=\sum_{i=k}^\infty x^i-\sum_{i=k+1}^\infty x^i=x^k\)
so that:
\(\sum_{i=k}^\infty x^i=\frac{x^k}{1-x}\)
It remains to substitute k=1 and \(x=\frac{5}{12}\)

0

Expert Community at Your Service

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

Relevant Questions

asked 2021-12-22
I have to calculate the series
\(\displaystyle{\sum_{{{k}={2}}}^{\infty}}{\left({\frac{{{1}}}{{{k}}}}-{\frac{{{1}}}{{{k}+{2}}}}\right)}\)
Using the definition:
\(\displaystyle{L}=\lim_{{{n}\to\infty}}{S}_{{n}}=\lim_{{{n}\to\infty}}{\sum_{{{k}={0}}}^{{n}}}{a}^{{k}}\)
Obviously \(\displaystyle\lim_{{{n}\to\infty}}{\left({\frac{{{1}}}{{{n}}}}-{\frac{{{1}}}{{{n}+{2}}}}\right)}={0}\), but I don't think that this is the right way to calculate the value of the series.
asked 2021-12-23
I was asked to compute this series:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{\left({2}{n}-{1}\right)}^{{2}}}}}\)
but by using the fact that \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}^{{2}}}}}={\frac{{\pi^{{2}}}}{{{6}}}}\)
asked 2022-01-02
How do I evaluate the power series
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}}}{{{9}^{{n}}}}}\)
without using the formula for infinite geometric series?
asked 2021-11-23
Find whether the series diverges and its sum:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\left(-{1}\right)}^{{{n}+{1}}}{\frac{{{3}}}{{{5}^{{n}}}}}\)
asked 2022-01-01
Looking to prove that the following series converges conditionally
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}{\left({1}+{n}\right)}^{{{\frac{{{1}}}{{{n}}}}}}}}{{{n}}}}\)
asked 2022-01-04
Consider the following infinite series:
\(\displaystyle{\sum_{{{k}={0}}}^{\infty}}{\frac{{{2}^{{k}}}}{{{1}+{\frac{{{1}}}{{{x}^{{{2}{k}}}}}}}}}\)
asked 2022-01-06
Using power series representation, calculate
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}{2}^{{n}}}}{{{3}^{{n}}}}}\)
...