Is this limit correct ?

[CristianTheodor]

 

[pka]

View attachment 12906

I cannot really read the attachment. But look at this.

 

[CristianTheodor]

 

[CristianTheodor]

I cannot really read the attachment. But look at this.

Yes , I'm wrong . But how to solve it ?

 

[pka]

Thank you for posting the clear image. But no your work is incorrect.
For the \(\displaystyle |x|<1\) we have \(\displaystyle \sum\limits_{k = 0}^\infty {(k + 1){x^k}} = \frac{1}{{{{(1 - x)}^2}}}\)
In your problem let \(\displaystyle x=\frac{1}{3}\)
To prove that sum start with \(\displaystyle S(x) = \sum\limits_{k = 0}^\infty {{x^k}} = \frac{1}{{1 - x}}\)
What sum is \(\displaystyle x\cdot S(x)~?\)

 

[ksdhart2]

I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:

\(\displaystyle \frac{n + 1}{3^n} \cdot 2 = \frac{n+2}{3^{n+1}}\)

But this is only true if \(n = -\frac{4}{5}\), so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for \(a_n\) that is completely incorrect and seems to appear from nowhere. The only thing this working shows correctly is taking the limit - If that were a valid expression for \(a_n\), then the limit would be 2.

If nothing else, we can see that the limit must be incorrect, as the \(a_n\) is a strictly increasing sequence (why?) and the terms already exceed 2 when \(n = 3\):

\(\displaystyle a_3 = \sum\limits_{k = 0}^{3} \frac{k+1}{3^k} = \frac{{\color{red}0} + 1}{3^{{\color{red}0}}} + \frac{{\color{red}1} + 1}{3^{{\color{red}1}}} + \frac{{\color{red}2} + 1}{3^{{\color{red}2}}} + \frac{{\color{red}3} + 1}{3^{{\color{red}3}}} = \frac{58}{27} > 2\)

To be honest, your bizarre and seemingly disjointed work makes me wonder if you're familiar with sigma notation. Is that the case? If it is, I've linked a page where you can refresh yourself on this notation.

 

[CristianTheodor]

I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:

\(\displaystyle \frac{n + 1}{3^n} \cdot 2 = \frac{n+2}{3^{n+1}}\)

But this is only true if \(n = -\frac{4}{5}\), so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for \(a_n\) that is completely incorrect and seems to appear from nowhere. The only thing this working shows correctly is taking the limit - If that were a valid expression for \(a_n\), then the limit would be 2.

If nothing else, we can see that the limit must be incorrect, as the \(a_n\) is a strictly increasing sequence (why?) and the terms already exceed 2 when \(n = 3\):

\(\displaystyle a_3 = \sum\limits_{k = 0}^{3} \frac{k+1}{3^k} = \frac{{\color{red}0} + 1}{3^{{\color{red}0}}} + \frac{{\color{red}1} + 1}{3^{{\color{red}1}}} + \frac{{\color{red}2} + 1}{3^{{\color{red}2}}} + \frac{{\color{red}3} + 1}{3^{{\color{red}3}}} = \frac{58}{27} > 2\)

To be honest, your bizarre and seemingly disjointed work makes me wonder if you're familiar with sigma notation. Is that the case? If it is, I've linked a page where you can refresh yourself on this notation.

I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:

\(\displaystyle \frac{n + 1}{3^n} \cdot 2 = \frac{n+2}{3^{n+1}}\)

But this is only true if \(n = -\frac{4}{5}\), so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for \(a_n\) that is completely incorrect and seems to appear from nowhere. The only thing this working shows correctly is taking the limit - If that were a valid expression for \(a_n\), then the limit would be 2.

If nothing else, we can see that the limit must be incorrect, as the \(a_n\) is a strictly increasing sequence (why?) and the terms already exceed 2 when \(n = 3\):

\(\displaystyle a_3 = \sum\limits_{k = 0}^{3} \frac{k+1}{3^k} = \frac{{\color{red}0} + 1}{3^{{\color{red}0}}} + \frac{{\color{red}1} + 1}{3^{{\color{red}1}}} + \frac{{\color{red}2} + 1}{3^{{\color{red}2}}} + \frac{{\color{red}3} + 1}{3^{{\color{red}3}}} = \frac{58}{27} > 2\)

To be honest, your bizarre and seemingly disjointed work makes me wonder if you're familiar with sigma notation. Is that the case? If it is, I've linked a page where you can refresh yourself on this notation.

That's not 2 , it's q . I use the geometric series sum . The q is the ratio .

 

[pka]

I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:

If \(\displaystyle |x|<1\) then this is a well known sum \(\displaystyle S(x) = \sum\limits_{k = 0}^\infty {{x^k}} = \frac{1}{{1 - x}}\)
We multiply by \(\displaystyle x\) in order to get \(\displaystyle k+1\) into the sum \(\displaystyle x \cdot S(x) = \sum\limits_{k = 0}^\infty {{x^{k + 1}}} = \frac{x}{{1 - x}}\)
Finding the derivative of that we get
\(\displaystyle S(x) + x \cdot S'(x) = \sum\limits_{k = 0}^\infty {(k + 1){x^k}} = \frac{1}{{{{(1 - x)}^2}}}\)
Let \(\displaystyle x=\frac{1}{3}\) or \(\displaystyle \sum\limits_{k = 0}^\infty {\frac{{k + 1}}{{{3^k}}}} = \frac{1}{{{{\left( {1 - \frac{1}{3}} \right)}^2}}} = \frac{9}{4}\)

 

[CristianTheodor]

If \(\displaystyle |x|<1\) then this is a well known sum \(\displaystyle S(x) = \sum\limits_{k = 0}^\infty {{x^k}} = \frac{1}{{1 - x}}\)
We multiply by \(\displaystyle x\) in order to get \(\displaystyle k+1\) into the sum \(\displaystyle x \cdot S(x) = \sum\limits_{k = 0}^\infty {{x^{k + 1}}} = \frac{x}{{1 - x}}\)
Finding the derivative of that we get
\(\displaystyle S(x) + x \cdot S'(x) = \sum\limits_{k = 0}^\infty {(k + 1){x^k}} = \frac{1}{{{{(1 - x)}^2}}}\)
Let \(\displaystyle x=\frac{1}{3}\) or \(\displaystyle \sum\limits_{k = 0}^\infty {\frac{{k + 1}}{{{3^k}}}} = \frac{1}{{{{\left( {1 - \frac{1}{3}} \right)}^2}}} = \frac{9}{4}\)

That's so clever.Thank you so much .

 

[Steven G]

Yes , I'm wrong . But how to solve it ?

Being right or wrong is NOT the real problem. You and PKA seem to be solving different problems! PKA is solving a summation (which I think that you want) WHILE you are solving a limit. Why is this?

 

[pka]

Being right or wrong is NOT the real problem. You and PKA seem to be solving different problems! PKA is solving a summation (which I think that you want) WHILE you are solving a limit. Why is this?

To Jomo, are you really serious?
Surely you know what it what it means for a infinite series to converse means? Don't you??
Suppose that \(\displaystyle {a_n} = \sum\limits_{k = 0}^n {f(k)} \) then the sequence \(\displaystyle a_n\) is a numerical sequence of partial sums.
If the sequence converges \(\displaystyle a_n\to S\) then we say \(\displaystyle \sum\limits_{k = 0}^\infty {f(k)} = S\).
Jomo, did you read the original postings? In it CristianTheodor asked about the partial sum \(\displaystyle {a_n} = \sum\limits_{k = 1}^n {\frac{{k + 1}}{{{3^k}}}} \)
Do you understand summations of infinite series? If not please don't say that I miss-read the question.
Look at reply #9 and then tell me that I am mistaken.

 

[Steven G]

To Jomo, are you really serious?
Surely you know what it what it means for a infinite series to converse means? Don't you??An infinite series converges if the infinite sum is finite
Suppose that \(\displaystyle {a_n} = \sum\limits_{k = 0}^n {f(k)} \) then the sequence \(\displaystyle a_n\) is a numerical sequence of partial sums. I understand that.
If the sequence converges \(\displaystyle a_n\to S\) then we say \(\displaystyle \sum\limits_{k = 0}^\infty {f(k)} = S\). Sure
Jomo, did you read the original postings? In it CristianTheodor asked about the partial sum \(\displaystyle {a_n} = \sum\limits_{k = 1}^n {\frac{{k + 1}}{{{3^k}}}} \) Yes he did
Do you understand summations of infinite series? If not please don't say that I miss-read the question.
Look at reply #9 and then tell me that I am mistaken. Never said you were mistaken

Read my bold comments above.
You got me good I admit that. At a quick glance the OP was looking at limits and you were finding an infinite sum. Not thinking I thought they were not the same thing. As I said you got me good!