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

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 $$\frac{23}{2}$$

### Relevant Questions

Write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges.
$$\sum_{n=2}^\infty\frac{1}{4^n}$$
Write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges.
$$\displaystyle{\sum_{{{n}={2}}}^{\infty}}{\frac{{{1}}}{{{4}^{{n}}}}}$$
Write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges.
$$\displaystyle{\sum_{{{n}={2}}}^{\infty}}{\frac{{{1}}}{{{4}^{{n}}}}}$$
Write out the first few terms of the series
$$\sum_{n=0}^\infty\frac{(-1)^n}{5^n}$$
What is the​ series' sum?
Find the sum of the series:
$$\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))$$
a) Find the Maclaurin series for the function
$$f(x)=\frac11+x$$
b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function
$$g(x)=\frac{1}{(x+1)^2}$$
c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function
$$h(x)=\frac{1}{(x+1)^3}$$
d) Find the sum of the series
$$\sum_{n=3}^\infty \frac{n(n-1)}{2n}$$
This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is $$\frac72$$
Consider the telescoping series
$$\sum_{k=3}^\infty(\sqrt{k}-\sqrt{k-2})$$
a) Apply the Divvergence Test to the series. Show all details. What conclusions, if any, can you make about the series?
b) Write out the partial sums $$s_3,s_4,s_5$$ and $$s_6$$.
c) Compute the n-th partial sum $$s_n$$, and put it in closed form.
$$\displaystyle{\sum_{{{k}={3}}}^{\infty}}{\left(\sqrt{{{k}}}-\sqrt{{{k}-{2}}}\right)}$$
b) Write out the partial sums $$\displaystyle{s}_{{3}},{s}_{{4}},{s}_{{5}}$$ and $$\displaystyle{s}_{{6}}$$.
c) Compute the n-th partial sum $$\displaystyle{s}_{{n}}$$, and put it in closed form.
$$\sum_{n=2}^\infty(-\frac{5}{3})^n(\frac{2}{5})^{n+1}$$
Determine if the series $$\sum_{n=0}^\infty a_n$$ is convergent or divergent if the partial sum of the n terms of the series is given below. If the series is convergent, determine the value of the series.
$$S_n=\frac{5+8n^2}{2-7n^2}$$