# Thread: An easy infinite series

1. ## An easy infinite series

I must look pathetic, but I'm beginning with infinite series.
I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
The series is : $\displaystyle \sum_{i=1}^{+\infty} \frac{i}{3^i}$.
I tried the ratio test, writing down $\displaystyle a_i$ and $\displaystyle a_{i+1}$, dividing $\displaystyle a_i$ by $\displaystyle a_{i+1}$ I finally got $\displaystyle \frac{3i}{i+1}$. Taking the limit when $\displaystyle i$ tends to $\displaystyle +\infty$ I got $\displaystyle 3$. As $\displaystyle 3>1$, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.

2. Originally Posted by arbolis
I must look pathetic, but I'm beginning with infinite series.
I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
The series is : $\displaystyle \sum_{i=1}^{+\infty} \frac{i}{3^i}$.
I tried the ratio test, writing down $\displaystyle a_i$ and $\displaystyle a_{i+1}$, dividing $\displaystyle a_i$ by $\displaystyle a_{i+1}$ I finally got $\displaystyle \frac{3i}{i+1}$. Taking the limit when $\displaystyle i$ tends to $\displaystyle +\infty$ I got $\displaystyle 3$. As $\displaystyle 3>1$, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
Root test

$\displaystyle \lim_{i\to\infty}\bigg(\frac{i}{3^i}\bigg)^{\frac{ 1}{i}}=\lim_{i\to\infty}\frac{i^{\frac{1}{i}}}{3}= \frac{1}{3}<1$

thus convergent

3. Originally Posted by arbolis
I must look pathetic, but I'm beginning with infinite series.
I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
The series is : $\displaystyle \sum_{i=1}^{+\infty} \frac{i}{3^i}$.
I tried the ratio test, writing down $\displaystyle a_i$ and $\displaystyle a_{i+1}$, dividing $\displaystyle a_i$ by $\displaystyle a_{i+1}$ I finally got $\displaystyle \frac{3i}{i+1}$. Taking the limit when $\displaystyle i$ tends to $\displaystyle +\infty$ I got $\displaystyle 3$. As $\displaystyle 3>1$, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
Also by ratio test

$\displaystyle \lim_{i\to\infty}\bigg|\frac{i+1}{3\cdot{3^i}}\cdo t\frac{3^i}{i}\bigg|=\lim_{i\to\infty}\bigg|\frac{ i+1}{3i}\bigg|=\frac{1}{3}$

Remember the ratio and root test always yield the same number

4. You can use the geometric series formula.

$\displaystyle \sum_{i=1}^{\infty}\frac{i}{3^{i}}=\frac{1}{2}\cdo t\frac{1}{1-\frac{1}{3}}=\frac{3}{4}$

5. Originally Posted by galactus
You can use the geometric series formula.

$\displaystyle \sum_{i=1}^{\infty}\frac{i}{3^{i}}=\frac{1}{2}\cdo t\frac{1}{1-\frac{1}{3}}=\frac{3}{4}$
But this is not geometric, it has an extra term in it. It happen to the the derivative of the geometric series.

6. Originally Posted by arbolis
I must look pathetic, but I'm beginning with infinite series.
I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
The series is : $\displaystyle \sum_{i=1}^{+\infty} \frac{i}{3^i}$.
I tried the ratio test, writing down $\displaystyle a_i$ and $\displaystyle a_{i+1}$, dividing $\displaystyle a_i$ by $\displaystyle a_{i+1}$ I finally got $\displaystyle \frac{3i}{i+1}$. Taking the limit when $\displaystyle i$ tends to $\displaystyle +\infty$ I got $\displaystyle 3$. As $\displaystyle 3>1$, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
Also

There $\displaystyle \exists{N}\backepsilon\forall{n>N}\text{ }\frac{n}{3^n}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\b igg)^n$

Thus convergent by comparison test

Originally Posted by arbolis
I must look pathetic, but I'm beginning with infinite series.
I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
The series is : $\displaystyle \sum_{i=1}^{+\infty} \frac{i}{3^i}$.
I tried the ratio test, writing down $\displaystyle a_i$ and $\displaystyle a_{i+1}$, dividing $\displaystyle a_i$ by $\displaystyle a_{i+1}$ I finally got $\displaystyle \frac{3i}{i+1}$. Taking the limit when $\displaystyle i$ tends to $\displaystyle +\infty$ I got $\displaystyle 3$. As $\displaystyle 3>1$, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
Also
since $\displaystyle a_{i+1}<a_i$ for a finite value of i and every term is positive
$\displaystyle \int_1^{\infty}\frac{x}{3^x}dx=.58$

thus the integral converges by the integral test

if you would like I can give you some practice problems and help you through them

Originally Posted by galactus
You can use the geometric series formula.

$\displaystyle \sum_{i=1}^{\infty}\frac{i}{3^{i}}=\frac{1}{2}\cdo t\frac{1}{1-\frac{1}{3}}=\frac{3}{4}$
This is equal to $\displaystyle x\cdot\frac{d}{dx}\bigg[\sum_{n=1}^{\infty}\frac{1}{3^x}\bigg]=\frac{x}{(1-x)^2}$

evaluating at 1/3 gives the correct result...which was 3/4

7. I just realized I made the ratio test but inversed! So I got $\displaystyle 3$ instead of $\displaystyle \frac{1}{3}$. I also notice that the series value is not equal to the result of the ratio test (yeah, why would it be so?).
From Mathstud :
Also
since $\displaystyle a_{i+1}<a_i$ for a finite value of i and every term is positive
$\displaystyle \int_1^{\infty}\frac{x}{3^x}dx=.58$
. Is this the value of the series? If yes, any series can be written as an improper integral and the value of the series is always equals to the improper integral?

To
if you would like I can give you some practice problems and help you through them
I think it's a good idea, but I'm not sure I will do a lot of them these days (since I have an exam coming very soon and it has almost nothing to see with infinite series). But why not doing a bit? So yes.
Finally I didn't understand well the geometric series formula. I will investigate about it.

8. Originally Posted by arbolis
I must look pathetic, but I'm beginning with infinite series.
I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
The series is : $\displaystyle \sum_{i=1}^{+\infty} \frac{i}{3^i}$.
I tried the ratio test, writing down $\displaystyle a_i$ and $\displaystyle a_{i+1}$, dividing $\displaystyle a_i$ by $\displaystyle a_{i+1}$ I finally got $\displaystyle \frac{3i}{i+1}$. Taking the limit when $\displaystyle i$ tends to $\displaystyle +\infty$ I got $\displaystyle 3$. As $\displaystyle 3>1$, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
$\displaystyle \sum_{i=1}^{\infty}{i}{x^i}=x \sum_{i=1}^{\infty}\frac{d}{dx}{x^i}=x \times \frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{x}{(1-x)^2}$

which all works for $\displaystyle |x|<1$, so put x=1/3, to get:

$\displaystyle \sum_{i=1}^{\infty}\frac{i}{3^i}=\frac{(1/3)}{(2/3)^2}=\frac{3}{4}$

RonL

9. Thanks CaptainBlack! Nicely explained. (Sorry if I didn't quote your answer, I have some problem with the quoting part).

10. Originally Posted by Mathstud28
Also

There $\displaystyle \exists{N}\backepsilon\forall{n>N}\text{ }\frac{n}{3^n}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\b igg)^n$

Thus convergent by comparison test

Also
since $\displaystyle a_{i+1}<a_i$ for a finite value of i and every term is positive
$\displaystyle \int_1^{\infty}\frac{x}{3^x}dx=.58$

thus the integral converges by the integral test

if you would like I can give you some practice problems and help you through them

This is equal to $\displaystyle x\cdot\frac{d}{dx}\bigg[\sum_{n=1}^{\infty}\frac{1}{3^x}\bigg]=\frac{x}{(1-x)^2}$

evaluating at 1/3 gives the correct result...which was 3/4
Yes, I reckon I should have explained more thoroughly. It was late and I was getting droopy.

11. $\displaystyle G(x)=\sum_{k=0}^{\infty}{a_k\cdot{x^k}}$ where we want $\displaystyle a_k=k$ (we'll assume $\displaystyle x$ will be on the interval of convergence, that is $\displaystyle (-1;1)$ )

Now the condition can be stated as:
$\displaystyle a_{k+1}=a_k+1$ where $\displaystyle a_0=0$

Multiplying by $\displaystyle x^{k+1}$ on both sides:
$\displaystyle a_{k+1}\cdot{x^{k+1}}=a_k\cdot{x^{k+1}}+x^{k+1}$
That is:
$\displaystyle a_{k+1}\cdot{x^{k+1}}=x\cdot{a_k\cdot{x^{k}}}+x^{k +1}$

Now sum from $\displaystyle k=0$ to $\displaystyle +\infty$:
$\displaystyle \sum_{k=0}^{\infty}{a_{k+1}\cdot{x^{k+1}}}=x\cdot{ \sum_{k=0}^{\infty}{a_{k}\cdot{x^{k}}}}+\sum_{k=0} ^{\infty}{x^{k+1}}$

Thus we have:
$\displaystyle G(x)-a_0=x\cdot{G(x)}+\sum_{k=1}^{\infty}{x^{k}}=x\cdot {G(x)}+\frac{x}{1-x}$

Thus, since $\displaystyle a_0=0$:
$\displaystyle (1-x)\cdot{G(x)}=\frac{x}{1-x}$ and $\displaystyle G(x)=\frac{x}{(1-x)^2}$

You can use this method to work with other sequences where you have a recurrence