# Thread: Examine the convergence of the series

1. ## Examine the convergence of the series

Question : Using the Ratio Test for the convergence of an infinite series, examine the convergence of the series

$\frac{1}{2} + \frac{1}{2.2^2}+\frac{1}{3.2^3}+ \frac{1}{4.2^4}+...$

Solution

My work::::::::::::

$u_n$ = $\sum_{n=1}^{ \infty} \ \frac{1}{n.2^n}$.............Is this correct?????

2. Originally Posted by zorro
Question : Using the Ratio Test for the convergence of an infinite series, examine the convergence of the series

$\frac{1}{2} + \frac{1}{2.2^2}+\frac{1}{3.2^3}+ \frac{1}{4.2^4}+...$

Solution

My work::::::::::::

$u_n$ = $\sum_{n=1}^{ \infty} \ \frac{1}{n.2^n}$.............Is this correct?????

$u_n=\frac{1}{n2^n}\Longrightarrow \frac{u_{n+1}}{u_n}=\frac{1}{2}\cdot \frac{n}{n+1} \xrightarrow [n\to\infty]{}\frac{1}{2}<1$

Tonio

3. Obviously $\frac{1}{n2^{n}}\le \frac{1}{2^{n}},$ so the general term of the series was also comparable to a geometric one.

4. Originally Posted by tonio
$u_n=\frac{1}{n2^n}\Longrightarrow \frac{u_{n+1}}{u_n}=\frac{1}{2}\cdot \frac{n}{n+1} \xrightarrow [n\to\infty]{}\frac{1}{2}<1$

Tonio
No tonio i meant the formation of the summation is that correct ??????

5. Originally Posted by zorro
No tonio i meant the formation of the summation is that correct ??????

If you're asking whether the series is what you wrote then the answer is yes, but I don't understand why you wrote $u_n$ ...
Perhaps you meant to write the n-th term of the sequence of partials sums of the series? If so then it is $u_n=\sum\limits_{k=1}^n\frac{1}{k2^k}$

Tonio

6. no i wanted to know if the $u_n$ which i have used is corect or know because only up on that the $u_{n+1}$ would be decided that is why wnated to know if the starting is correct or no

7. Originally Posted by zorro
no i wanted to know if the $u_n$ which i have used is corect or know because only up on that the $u_{n+1}$ would be decided that is why wnated to know if the starting is correct or no

Ok, I don't understand what you're asking, but your $u_n=\sum\limits_{n=1}^\infty\frac{1}{n2^n}$ is the WHOLE series...! It makes no sense.

Tonio

8. Originally Posted by zorro
Question : Using the Ratio Test for the convergence of an infinite series, examine the convergence of the series

$\frac{1}{2} + \frac{1}{2.2^2}+\frac{1}{3.2^3}+ \frac{1}{4.2^4}+...$

Solution

My work::::::::::::

$u_n$ = $\sum_{n=1}^{ \infty} \ \frac{1}{n.2^n}$.............Is this correct?????
Learning to use "Ratio Test for the convergence" is important. But sometimes you can "eyeball" the answer by looking at the form of the series. I generally enjoy summing up such series. I am sorry for swaying from the main topic, but I want to demonstrate this method. It wont work everywhere but with experience you will know where to use it

The 'n' factor in the denominator looks as though it $x^{n-1}$ has been integrated. It is immediate that

for $|x| < 1$,
$\frac1{1-y} = \sum_{n=1}^{ \infty} y^{n-1}$
$\int_0^x \frac1{1-y}\, dy = \sum_{n=1}^{ \infty} \int_0^x y^{n-1}\, dy$
$-\ln(1-y)\bigg{|}_0^x = \sum_{n=1}^{ \infty} \frac{y^n}{n}\bigg{|}_0^x$
$-\ln(1-x) = \sum_{n=1}^{ \infty} \frac{x^n}{n}$
Now substitute $x = \frac12$ and so you even know that $\sum_{n=1}^{ \infty} \ \frac{1}{n.2^n} = \ln 2$

9. Thanks mite