Find the sum of the series $\displaystyle \sum_{n=0}^{\infty}$$\displaystyle \frac{3n+5}{4n} I know that the limit of the function is also the sum of the series but I got 0 for the limit which I'm not so sure if it is right can anyone help me and see if thats correct? 2. ## Re: Sum of Series Originally Posted by JC05 Find the sum of the series \displaystyle \sum_{n=0}^{\infty}$$\displaystyle \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 \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.

3. ## Re: Sum of Series

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

4. ## Re: Sum of Series

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.

5. ## Re: Sum of Series

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

work from there.

6. ## Re: Sum of Series

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 \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}, he wrote \displaystyle \displaystyle \begin{align*} \frac{3n + 5}{4n} \end{align*}, so your simplification is not valid.

7. ## Re: Sum of Series

Originally Posted by Prove It
Don't confuse the OP. He did not write \displaystyle \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}, he wrote \displaystyle \displaystyle \begin{align*} \frac{3n + 5}{4n} \end{align*}, so your simplification is not valid.
MY BAD!!! it is \displaystyle \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}

No wonder I was getting super confused ...

8. ## Re: Sum of Series

Originally Posted by JC05
MY BAD!!! it is \displaystyle \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 \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 \displaystyle \begin{align*} a = 1, r = \frac{3}{4} \end{align*} and the second is an infinite geometric series with \displaystyle \displaystyle \begin{align*} a = 1, r = \frac{1}{4} \end{align*}.

9. ## Re: Sum of Series

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

10. ## Re: Sum of Series

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

11. ## Re: Sum of Series

Originally Posted by Prove It
Why...?
Ok so the first thing i did was calculate $\displaystyle \frac{1}{1-\frac{3}{4}}$ which equal 4

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

Then i combined both equation to get 10.33

12. ## Re: Sum of Series

Originally Posted by JC05
Why...?
Ok so the first thing i did was calculate $\displaystyle \frac{1}{1-\frac{3}{4}}$ which equal 4

then i did the second one $\displaystyle \frac{1}{1-\frac{1}{4}}$ 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.

13. ## Re: Sum of Series

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

14. ## Re: Sum of Series

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 \displaystyle \begin{align*} \frac{32}{3} \end{align*}.