Convergent Geometric Series

[urimagic]

Hi friends,

I trust you are all well. I have tried another Sigma related problem but I'm afraid I'm not too sure about what and how I did to solve the sum. Please coul;d someone have a look and correct me where I've gone wrong?..I honestly will appreciate any help I can get.

 

[Harry_the_cat]

You have found r in terms of a. The question asks for a in terms of r.

Also, is it n=3 under the sigma sign in the question? If so, Your second line in (2) is incorrect. If the sum equals 1/4, then the same sum plus 2 can't also equal 1/4.

 

[pka]

Why do you not post something that is readable?
If the question is about geometric series are they finite or infinite?
If the common ratio $r\ne 0$ then $\sum\limits_{n = 1}^N {a{r^{n - 1}}} = a\frac{{(1 - {r^N})}}{{(1 - r)}}$ SEE HERE

$$$$

 

[mmm4444bot]

Why do you not post something that is readable?

I can read it. What parts do you need help with? ?
$\;$

 

[pka]

I can read it. What parts do you need help with? ?
$\;$

Good you. I can read none of it. Any student of mine would have to redo it.

 

[urimagic]

You have found r in terms of a. The question asks for a in terms of r.

Also, is it n=3 under the sigma sign in the question? If so, Your second line in (2) is incorrect. If the sum equals 1/4, then the same sum plus 2 can't also equal 1/4.

Hello,

oops yes, I see that now...Thank you

 

[urimagic]

Why do you not post something that is readable?
If the question is about geometric series are they finite or infinite?
If the common ratio $r\ne 0$ then $\sum\limits_{n = 1}^N {a{r^{n - 1}}} = a\frac{{(1 - {r^N})}}{{(1 - r)}}$ SEE HERE

$$$$

Hi pka,

sorry about that, I'll fix it..

 

[urimagic]

I'll retry the problem first, and if stuck, will post again...Thank you for your responses..

 

[urimagic]

Ok, I tried, but obviously with my limited understanding of this, I ran into a problem...
Please help, why is this not working out?

 

[BigBeachBanana]

Ok, I tried, but obviously with my limited understanding of this, I ran into a problem...View attachment 35484
Please help, why is this not working out?

$$\begin{aligned} \sum_{n=1}^\infty T_n = \dfrac{a}{1-r}\\ T_1 + T_2 + \sum_{n=3}^\infty T_n = \dfrac{a}{1-r}\\ 2 + \dfrac{1}{4} = \dfrac{a}{1-r} \end{aligned}$$Use the equation from (1) to solve for a and r simultaneously.

 

[urimagic]

$$\begin{aligned} \sum_{n=1}^\infty T_n = \dfrac{a}{1-r}\\ T_1 + T_2 + \sum_{n=3}^\infty T_n = \dfrac{a}{1-r}\\ 2 + \dfrac{1}{4} = \dfrac{a}{1-r} \end{aligned}$$Use the equation from (1) to solve for a and r simultaneously.

Thank you, I will tackle that soon...appreciate..

 

[urimagic]

Thank you, I will tackle that soon...appreciate..

I think I got it...Calculating the 2nd term using r=1/3, and a=3/2, I get 1/2, which means t1+t2=2.. Thank you BigBeachBanana. Just one thing if you do not mind:..how do I word the sum...meaning..is it right to say The sum from the 3rd term to infinity is 1/4, OR, the sum from the 3rd term to the nth term is 1/4...why I'm asking is that Tn=1/4, but the 1/4 is replaced with the formula a/1-r..and then the 1/4 is treated as a constant and added to terms 1 and 2?..I'm not sure if I'm making any sense as to what I'm asking you to please explain to me?

 

[BigBeachBanana]

I think I got it...Calculating the 2nd term using r=1/3, and a=3/2, I get 1/2, which means t1+t2=2.. Thank you BigBeachBanana. Just one thing if you do not mind:..how do I word the sum...meaning..is it right to say The sum from the 3rd term to infinity is 1/4, OR, the sum from the 3rd term to the nth term is 1/4...why I'm asking is that Tn=1/4, but the 1/4 is replaced with the formula a/1-r..and then the 1/4 is treated as a constant and added to terms 1 and 2?..I'm not sure if I'm making any sense as to what I'm asking you to please explain to me?View attachment 35489

The difference is between a finite vs an infinite sum.
For a finite sum, the upper limit is some integer $N$ i.e. $$\sum_{n=1}^N T_n = \dfrac{a(1-r^N)}{1-r}$$In your case, your upper limit is infinity so we're talking about an infinite sum here i.e.
$$\lim_{N \to \infty} \sum_{n=1}^N T_n =\lim_{N \to \infty} \dfrac{a(1-r^N)}{1-r} = \dfrac{a}{1-r}$$
Does this answer your question?

 

[urimagic]

The difference is between a finite vs an infinite sum.
For a finite sum, the upper limit is some integer $N$ i.e. $$\sum_{n=1}^N T_n = \dfrac{a(1-r^N)}{1-r}$$In your case, your upper limit is infinity so we're talking about an infinite sum here i.e.
$$\lim_{N \to \infty} \sum_{n=1}^N T_n =\lim_{N \to \infty} \dfrac{a(1-r^N)}{1-r} = \dfrac{a}{1-r}$$
Does this answer your question?

Hi, not entirely, I have actually made a calculation to assist with my question...We know T1 = 3/2, r = 1/3, so T2 = 3/6, (1/2), which gives the sum of T1+T2 = 2...now the value of T3 = 3/18 (1/6), , so I see the sum of the first 3 terms = 2,17, so I'm seeing two things here: (1) The sum value of all terms as we add them to infinity, moves further away from 1/4 and (2), The actual term values also get smaller as we add them...So I do not see how this 1/4 plays any true role here, because going to infinity, we never get closer to it, ...not sum value wise, nor term value wise..so, what is this Tn = 1/4.?..I guess what I'm getting at is that we are not moving closer to 1/4, but instead further away, whichever way we look at it...

 

[BigBeachBanana]

Hi, not entirely, I have actually made a calculation to assist with my question...We know T1 = 3/2, r = 1/3, so T2 = 3/6, (1/2), which gives the sum of T1+T2 = 2...now the value of T3 = 3/18 (1/6), , so I see the sum of the first 3 terms = 2,17, so I'm seeing two things here: (1) The sum value of all terms as we add them to infinity, moves further away from 1/4 and (2), The actual term values also get smaller as we add them...So I do not see how this 1/4 plays any true role here, because going to infinity, we never get closer to it, ...not sum value wise, nor term value wise..so, what is this Tn = 1/4.?..I guess what I'm getting at is that we are not moving closer to 1/4, but instead further away, whichever way we look at it...

Pay attention to the index.
$$\sum_{\red{n=1}}^\infty T_n = \dfrac{9}{4}\\ \sum_{\red{n=3}}^\infty T_n = \dfrac{1}{4}\\$$It will approach 1/4 if you start the sum from the 3rd term onwards and 9/4 if you start the sum from the first term.

 

[urimagic]

Pay attention to the index.
$$\sum_{\red{n=1}}^\infty T_n = \dfrac{9}{4}\\ \sum_{\red{n=3}}^\infty T_n = \dfrac{1}{4}\\$$It will approach 1/4 if you start the sum from the 3rd term onwards and 9/4 if you start the sum from the first term.

Thanks a lot, have a great day..