Write out the first few terms of each series to show how the series start. Then find out the sum of the series. sum_{n=0}^infty(frac{5}{2^n}+frac{1}{3^n})

Question
Series
asked 2021-03-08
Write out the first few terms of each series to show how the series start. Then find out the sum of the series.
\(\sum_{n=0}^\infty(\frac{5}{2^n}+\frac{1}{3^n})\)

Answers (1)

2021-03-09
Consider the given series as
\(I=\sum_{n=0}^\infty(\frac{5}{2^n}+\frac{1}{3^n})\)
To find the few terms ,substitute the value of n from 0 to infinity in given series i.e.
\(I=(\frac{5}{2^0}+\frac{1}{3^0})+(\frac{5}{2^1}+\frac{1}{3^1})+(\frac{5}{2^2}+\frac{1}{3^2})+...\)
\(=(5+1)+(\frac{15+2}{6})+(\frac{45+4}{36})+...\)
\(=6+\frac{17}{6}+\frac{49}{36}+...\)
Since the given series is geometric series. So, use formula of geometric sum of infinite series
\(S_n=\frac{a(1-r^n)}{1-r}\)
where a is the first term of series and r is geometric ratio between every two terms.
As, the given series can be written as
\(I=\sum_{n=0}^\infty(\frac{5}{2^n})+\sum_{n=0}^\infty(\frac{1}{3^n})\)
If the series
\(\sum a_n=\sum_{n=0}^\infty(\frac{5}{2^n})\quad\text{and}\quad\sum b_n=\sum_{n=0}^\infty\frac{1}{3^n}\)
are convergent series then it can be written in the form of
\(\sum a_n+b_n=\sum a_n+\sum b_n\)
Check the convergence of the series by ratio test
\(\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}\frac{\frac{5}{2^{n+1}}}{\frac{5}{2^n}}\)
\(=\lim_{n\rightarrow\infty}\frac{2^n}{2^{n+1}}\)
\(=\lim_{n\rightarrow\infty}\frac{2^n}{2^n(2)}\)
\(=\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\frac{1}{2}<1\)</span>
As, the limit for above series is less than 1 . Therefore, the above series is convergent.
Now, check for another series
\(=\lim_{n\rightarrow\infty}\frac{b_{n+1}}{b_n}=\frac{\frac{1}{3^{n+1}}}{\frac{1}{3^n}}\)
\(=\lim_{n\rightarrow\infty}\frac{3^n}{3^n(3)}\)
\(=\lim_{n\rightarrow\infty}\frac{b_{n+1}}{b_n}=\frac{1}{3}<1\)</span>
It implies that the limit of above series is less than 1 and hence is convergent.
Therefore, the given series is
\(I=5\sum_{n=0}^\infty(\frac{1}{2^n})+\sum_{n=0}^{\infty}(\frac{1}{3^n})\)
By using formula of sum of infinite series
\(5(2)+\frac{3}{2}\)
\(\frac{23}{2}\)
Hence,the sum of given infinite series is \(\frac{23}{2}\)
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