Is this limit correct ?

Thank you for posting the clear image. But no your work is incorrect.
For the x<1\displaystyle |x|<1 we have k=0(k+1)xk=1(1x)2\displaystyle \sum\limits_{k = 0}^\infty {(k + 1){x^k}} = \frac{1}{{{{(1 - x)}^2}}}
In your problem let x=13\displaystyle x=\frac{1}{3}
To prove that sum start with S(x)=k=0xk=11x\displaystyle S(x) = \sum\limits_{k = 0}^\infty {{x^k}} = \frac{1}{{1 - x}}
What sum is xS(x) ?\displaystyle x\cdot S(x)~?
 
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:

n+13n2=n+23n+1\displaystyle \frac{n + 1}{3^n} \cdot 2 = \frac{n+2}{3^{n+1}}

But this is only true if n=45n = -\frac{4}{5}, so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for ana_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 ana_n, then the limit would be 2.

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

a3=k=03k+13k=0+130+1+131+2+132+3+133=5827>2\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:

n+13n2=n+23n+1\displaystyle \frac{n + 1}{3^n} \cdot 2 = \frac{n+2}{3^{n+1}}

But this is only true if n=45n = -\frac{4}{5}, so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for ana_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 ana_n, then the limit would be 2.

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

a3=k=03k+13k=0+130+1+131+2+132+3+133=5827>2\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:

n+13n2=n+23n+1\displaystyle \frac{n + 1}{3^n} \cdot 2 = \frac{n+2}{3^{n+1}}

But this is only true if n=45n = -\frac{4}{5}, so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for ana_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 ana_n, then the limit would be 2.

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

a3=k=03k+13k=0+130+1+131+2+132+3+133=5827>2\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 .
 
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 x<1\displaystyle |x|<1 then this is a well known sum S(x)=k=0xk=11x\displaystyle S(x) = \sum\limits_{k = 0}^\infty {{x^k}} = \frac{1}{{1 - x}}
We multiply by x\displaystyle x in order to get k+1\displaystyle k+1 into the sum xS(x)=k=0xk+1=x1x\displaystyle x \cdot S(x) = \sum\limits_{k = 0}^\infty {{x^{k + 1}}} = \frac{x}{{1 - x}}
Finding the derivative of that we get
S(x)+xS(x)=k=0(k+1)xk=1(1x)2\displaystyle S(x) + x \cdot S'(x) = \sum\limits_{k = 0}^\infty {(k + 1){x^k}} = \frac{1}{{{{(1 - x)}^2}}}
Let x=13\displaystyle x=\frac{1}{3} or k=0k+13k=1(113)2=94\displaystyle \sum\limits_{k = 0}^\infty {\frac{{k + 1}}{{{3^k}}}} = \frac{1}{{{{\left( {1 - \frac{1}{3}} \right)}^2}}} = \frac{9}{4}
 
If x<1\displaystyle |x|<1 then this is a well known sum S(x)=k=0xk=11x\displaystyle S(x) = \sum\limits_{k = 0}^\infty {{x^k}} = \frac{1}{{1 - x}}
We multiply by x\displaystyle x in order to get k+1\displaystyle k+1 into the sum xS(x)=k=0xk+1=x1x\displaystyle x \cdot S(x) = \sum\limits_{k = 0}^\infty {{x^{k + 1}}} = \frac{x}{{1 - x}}
Finding the derivative of that we get
S(x)+xS(x)=k=0(k+1)xk=1(1x)2\displaystyle S(x) + x \cdot S'(x) = \sum\limits_{k = 0}^\infty {(k + 1){x^k}} = \frac{1}{{{{(1 - x)}^2}}}
Let x=13\displaystyle x=\frac{1}{3} or k=0k+13k=1(113)2=94\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 .
 
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?
 
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 an=k=0nf(k)\displaystyle {a_n} = \sum\limits_{k = 0}^n {f(k)} then the sequence an\displaystyle a_n is a numerical sequence of partial sums.
If the sequence converges anS\displaystyle a_n\to S then we say k=0f(k)=S\displaystyle \sum\limits_{k = 0}^\infty {f(k)} = S.
Jomo, did you read the original postings? In it CristianTheodor asked about the partial sum an=k=1nk+13k\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.

 
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 an=k=0nf(k)\displaystyle {a_n} = \sum\limits_{k = 0}^n {f(k)} then the sequence an\displaystyle a_n is a numerical sequence of partial sums. I understand that.
If the sequence converges anS\displaystyle a_n\to S then we say k=0f(k)=S\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 an=k=1nk+13k\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!
 
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