Sum of Series

**JC05** · April 28th 2012, 10:12 PM

Find the sum of the series $\sum_{n=0}^{\infty}$ $\frac{3n+5}{4n}$

I know that the limit of the function is also the sum of the series but I got 0 for the limit which I'm not so sure if it is right
can anyone help me and see if thats correct?

**Prove It** · April 28th 2012, 11:29 PM

Originally Posted by JC05

Find the sum of the series $\sum_{n=0}^{\infty}$ $\frac{3n+5}{4n}$

I know that the limit of the function is also the sum of the series but I got 0 for the limit which I'm not so sure if it is right
can anyone help me and see if thats correct?

No, the limit of the function is NOT the value of the sum. The reason you need to work out this limit is because a series will never be convergent if the limit of the function is not 0.

It's pretty easy to see that

$\displaystyle \begin{align*} \lim_{n \to \infty}\frac{3n + 5}{4n} &= \lim_{n \to \infty}\left(\frac{3}{4} + \frac{5}{4n}\right) \\ &= \frac{3}{4} \\ &\neq 0 \end{align*}$

The series is divergent.

**JC05** · April 28th 2012, 11:35 PM

So is the sum of the series 3/4
I kinda feel so stupid now ... I should've seen that
Thanks though!!!

**Prove It** · April 28th 2012, 11:45 PM

Originally Posted by JC05

So is the sum of the series 3/4
I kinda feel so stupid now ... I should've seen that
Thanks though!!!

No, the sum of the series is NOT 3/4. Since this limit is not 0, the series is DIVERGENT! It shoots off to infinity.

**manhole** · April 29th 2012, 11:01 AM

inf-sigma-0 3^n/4^n + inf-sigma-0 5/4^n

work from there.

**Prove It** · April 29th 2012, 07:26 PM

Originally Posted by manhole

inf-sigma-0 3^n/4^n + inf-sigma-0 5/4^n

work from there.

Don't confuse the OP. He did not write $\displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}$ , he wrote $\displaystyle \begin{align*} \frac{3n + 5}{4n} \end{align*}$ , so your simplification is not valid.

**JC05** · April 29th 2012, 07:36 PM

Originally Posted by Prove It

Don't confuse the OP. He did not write $\displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}$ , he wrote $\displaystyle \begin{align*} \frac{3n + 5}{4n} \end{align*}$ , so your simplification is not valid.

MY BAD!!! it is $\displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}$

No wonder I was getting super confused ...

**Prove It** · April 29th 2012, 07:40 PM

Originally Posted by JC05

MY BAD!!! it is $\displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}$

No wonder I was getting super confused ...

Well then do as you were instructed by by Manhole, his psychic powers must be a little more powerful and finely tuned than mine...

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}\frac{3^n + 5}{4^n} &= \sum_{n = 0}^{\infty}\left(\frac{3^n}{4^n} + \frac{5}{4^n}\right) \\ &= \sum_{n = 0}^{\infty}\left(\frac{3}{4}\right)^n + 5\sum_{n = 0}^{\infty}\left(\frac{1}{4}\right)^n \end{align*}$

The first is an infinite geometric series with $\displaystyle \begin{align*} a = 1, r = \frac{3}{4} \end{align*}$ and the second is an infinite geometric series with $\displaystyle \begin{align*} a = 1, r = \frac{1}{4} \end{align*}$ .

**JC05** · April 29th 2012, 07:48 PM

Originally Posted by Prove It

Well then do as you were instructed by by Manhole, his psychic powers must be a little more powerful and finely tuned than mine...

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}\frac{3^n + 5}{4^n} &= \sum_{n = 0}^{\infty}\left(\frac{3^n}{4^n} + \frac{5}{4^n}\right) \\ &= \sum_{n = 0}^{\infty}\left(\frac{3}{4}\right)^n + 5\sum_{n = 0}^{\infty}\left(\frac{1}{4}\right)^n \end{align*}$

The first is an infinite geometric series with $\displaystyle \begin{align*} a = 1, r = \frac{3}{4} \end{align*}$ and the second is an infinite geometric series with $\displaystyle \begin{align*} a = 1, r = \frac{1}{4} \end{align*}$ .

Its ok.. apparently manhole is in my class
So i think i solved the equation which came out to be 10.33

**Prove It** · April 29th 2012, 08:33 PM

Originally Posted by JC05

Its ok.. apparently manhole is in my class
So i think i solved the equation which came out to be 10.33

I disagree with your answer.

**JC05** · April 29th 2012, 08:42 PM

Originally Posted by Prove It

I disagree with your answer.

Why...?
Ok so the first thing i did was calculate $\frac{1}{1-\frac{3}{4}}$ which equal 4

then i did the second one $\frac{1}{1-\frac{1}{4}}$ which equal 1.33 and then i added 5 to it which makes 6.33

Then i combined both equation to get 10.33

**Prove It** · April 29th 2012, 09:11 PM

Originally Posted by JC05

Why...?
Ok so the first thing i did was calculate $\frac{1}{1-\frac{3}{4}}$ which equal 4

then i did the second one $\frac{1}{1-\frac{1}{4}}$ which equal 1.33 and then i added 5 to it which makes 6.33

Then i combined both equation to get 10.33

You're not supposed to add 5 to the second one, it's supposed to be MULTIPLIED by 5.

**JC05** · April 29th 2012, 09:23 PM

Originally Posted by Prove It

You're not supposed to add 5 to the second one, it's supposed to be MULTIPLIED by 5.

Sorry....

10.67

**Prove It** · April 30th 2012, 03:05 AM

Originally Posted by JC05

Sorry....

10.67

That's better.

Of course, we should always stay exact where possible, so it would be better if you keep the answer as $\displaystyle \begin{align*} \frac{32}{3} \end{align*}$ .

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