Find $sum_{n=1}^{infty} frac{x^{n}}{(1+x)(1+x^{2}) dots (1+x^{n})}$ [closed]
For $x>1$ find $sum_{n=1}^{infty} frac{x^n}{(1+x)(1+x^{2}) dots (1+x^{n})}$
I don't have any idea how to even initiate. Please suggest, how to begin.
real-analysis limits
asked Dec 27 '18 at 14:54
Mathsaddict
2458
closed as off-topic by abiessu, RRL, mrtaurho, amWhy, José Carlos Santos Dec 27 '18 at 19:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, RRL, mrtaurho, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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For $x>1$ find $sum_{n=1}^{infty} frac{x^n}{(1+x)(1+x^{2}) dots (1+x^{n})}$
I don't have any idea how to even initiate. Please suggest, how to begin.
real-analysis limits
asked Dec 27 '18 at 14:54
Mathsaddict
2458
closed as off-topic by abiessu, RRL, mrtaurho, amWhy, José Carlos Santos Dec 27 '18 at 19:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, RRL, mrtaurho, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Try adding up the first few terms to see if a pattern emerges...
– abiessu
Dec 27 '18 at 14:57
@Henry Lee I can't proceed after doing what you explained.
– Mathsaddict
Dec 27 '18 at 15:24
7
Write the numerator as $x^n = (1+x^n)-1$. You should get a telescopic sum.
– D. Thomine
Dec 27 '18 at 15:25
similar : math.stackexchange.com/q/255483
– Jean Marie
Dec 27 '18 at 15:55
@D.Thomine Thanks! I got my telescoping series.
– Mathsaddict
Dec 27 '18 at 16:35
add a comment |
For $x>1$ find $sum_{n=1}^{infty} frac{x^n}{(1+x)(1+x^{2}) dots (1+x^{n})}$
I don't have any idea how to even initiate. Please suggest, how to begin.
real-analysis limits
asked Dec 27 '18 at 14:54
Mathsaddict
2458
For $x>1$ find $sum_{n=1}^{infty} frac{x^n}{(1+x)(1+x^{2}) dots (1+x^{n})}$
I don't have any idea how to even initiate. Please suggest, how to begin.
real-analysis limits
real-analysis limits
asked Dec 27 '18 at 14:54
Mathsaddict
2458
asked Dec 27 '18 at 14:54
Mathsaddict
2458
asked Dec 27 '18 at 14:54
Mathsaddict
2458
asked Dec 27 '18 at 14:54
Mathsaddict
2458
asked Dec 27 '18 at 14:54
Mathsaddict
2458
2458
closed as off-topic by abiessu, RRL, mrtaurho, amWhy, José Carlos Santos Dec 27 '18 at 19:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, RRL, mrtaurho, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by abiessu, RRL, mrtaurho, amWhy, José Carlos Santos Dec 27 '18 at 19:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, RRL, mrtaurho, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Try adding up the first few terms to see if a pattern emerges...
– abiessu
Dec 27 '18 at 14:57
@Henry Lee I can't proceed after doing what you explained.
– Mathsaddict
Dec 27 '18 at 15:24
7
Write the numerator as $x^n = (1+x^n)-1$. You should get a telescopic sum.
– D. Thomine
Dec 27 '18 at 15:25
similar : math.stackexchange.com/q/255483
– Jean Marie
Dec 27 '18 at 15:55
@D.Thomine Thanks! I got my telescoping series.
– Mathsaddict
Dec 27 '18 at 16:35
add a comment |
Try adding up the first few terms to see if a pattern emerges...
– abiessu
Dec 27 '18 at 14:57
@Henry Lee I can't proceed after doing what you explained.
– Mathsaddict
Dec 27 '18 at 15:24
7
Write the numerator as $x^n = (1+x^n)-1$. You should get a telescopic sum.
– D. Thomine
Dec 27 '18 at 15:25
similar : math.stackexchange.com/q/255483
– Jean Marie
Dec 27 '18 at 15:55
@D.Thomine Thanks! I got my telescoping series.
– Mathsaddict
Dec 27 '18 at 16:35
Try adding up the first few terms to see if a pattern emerges...
– abiessu
Dec 27 '18 at 14:57
Try adding up the first few terms to see if a pattern emerges...
– abiessu
Dec 27 '18 at 14:57
@Henry Lee I can't proceed after doing what you explained.
– Mathsaddict
Dec 27 '18 at 15:24
@Henry Lee I can't proceed after doing what you explained.
– Mathsaddict
Dec 27 '18 at 15:24
7
7
Write the numerator as $x^n = (1+x^n)-1$. You should get a telescopic sum.
– D. Thomine
Dec 27 '18 at 15:25
Write the numerator as $x^n = (1+x^n)-1$. You should get a telescopic sum.
– D. Thomine
Dec 27 '18 at 15:25
similar : math.stackexchange.com/q/255483
– Jean Marie
Dec 27 '18 at 15:55
similar : math.stackexchange.com/q/255483
– Jean Marie
Dec 27 '18 at 15:55
@D.Thomine Thanks! I got my telescoping series.
– Mathsaddict
Dec 27 '18 at 16:35
@D.Thomine Thanks! I got my telescoping series.
– Mathsaddict
Dec 27 '18 at 16:35
add a comment |
1 Answer
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Prove by induction that the $k$th partial sum is $1-frac{x^k}{(1+x)cdots (1+x^k)}$. The limit is $1$; the error terms $to 0$ because the ratio of consecutive error terms is $frac{x}{1+x^k}$, which has a $ktoinfty$ limit of modulus $<1$ for $xne -1$.
answered Dec 27 '18 at 16:14
J.G.
23k22137
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Prove by induction that the $k$th partial sum is $1-frac{x^k}{(1+x)cdots (1+x^k)}$. The limit is $1$; the error terms $to 0$ because the ratio of consecutive error terms is $frac{x}{1+x^k}$, which has a $ktoinfty$ limit of modulus $<1$ for $xne -1$.
answered Dec 27 '18 at 16:14
J.G.
23k22137
add a comment |
Prove by induction that the $k$th partial sum is $1-frac{x^k}{(1+x)cdots (1+x^k)}$. The limit is $1$; the error terms $to 0$ because the ratio of consecutive error terms is $frac{x}{1+x^k}$, which has a $ktoinfty$ limit of modulus $<1$ for $xne -1$.
answered Dec 27 '18 at 16:14
J.G.
23k22137
add a comment |
Prove by induction that the $k$th partial sum is $1-frac{x^k}{(1+x)cdots (1+x^k)}$. The limit is $1$; the error terms $to 0$ because the ratio of consecutive error terms is $frac{x}{1+x^k}$, which has a $ktoinfty$ limit of modulus $<1$ for $xne -1$.
answered Dec 27 '18 at 16:14
J.G.
23k22137
Prove by induction that the $k$th partial sum is $1-frac{x^k}{(1+x)cdots (1+x^k)}$. The limit is $1$; the error terms $to 0$ because the ratio of consecutive error terms is $frac{x}{1+x^k}$, which has a $ktoinfty$ limit of modulus $<1$ for $xne -1$.
answered Dec 27 '18 at 16:14
J.G.
23k22137
edited Dec 27 '18 at 16:30
answered Dec 27 '18 at 16:14
J.G.
23k22137
answered Dec 27 '18 at 16:14
J.G.
23k22137
answered Dec 27 '18 at 16:14
J.G.
23k22137
23k22137
add a comment |
add a comment |
Try adding up the first few terms to see if a pattern emerges...
– abiessu
Dec 27 '18 at 14:57
@Henry Lee I can't proceed after doing what you explained.
– Mathsaddict
Dec 27 '18 at 15:24
7
Write the numerator as $x^n = (1+x^n)-1$. You should get a telescopic sum.
– D. Thomine
Dec 27 '18 at 15:25
similar : math.stackexchange.com/q/255483
– Jean Marie
Dec 27 '18 at 15:55
@D.Thomine Thanks! I got my telescoping series.
– Mathsaddict
Dec 27 '18 at 16:35