Proving $sum_{k = 0}^{infty} frac{1}{1 + |x|^{k}}$ converges if and only if $|x| > 1$
I would like to show
$$sum_{k = 0}^{infty}frac{1}{1 + |x|^{k}} $$
converges if and only if $|x| > 1$. I think that the best way to show the backwards direction is to assume we havve $|x| leq 1$ then maybe doing the integral test? But I didn't get anywhere with this. I don't relaly have an idea of how to do it with the other direction either.
Any help is appreciated.
calculus sequences-and-series convergence
edited Dec 10 '18 at 2:10
user587192
1,775214
asked Dec 10 '18 at 1:50
joseph
4329
add a comment |
I would like to show
$$sum_{k = 0}^{infty}frac{1}{1 + |x|^{k}} $$
converges if and only if $|x| > 1$. I think that the best way to show the backwards direction is to assume we havve $|x| leq 1$ then maybe doing the integral test? But I didn't get anywhere with this. I don't relaly have an idea of how to do it with the other direction either.
Any help is appreciated.
calculus sequences-and-series convergence
edited Dec 10 '18 at 2:10
user587192
1,775214
asked Dec 10 '18 at 1:50
joseph
4329
add a comment |
I would like to show
$$sum_{k = 0}^{infty}frac{1}{1 + |x|^{k}} $$
converges if and only if $|x| > 1$. I think that the best way to show the backwards direction is to assume we havve $|x| leq 1$ then maybe doing the integral test? But I didn't get anywhere with this. I don't relaly have an idea of how to do it with the other direction either.
Any help is appreciated.
calculus sequences-and-series convergence
edited Dec 10 '18 at 2:10
user587192
1,775214
asked Dec 10 '18 at 1:50
joseph
4329
I would like to show
$$sum_{k = 0}^{infty}frac{1}{1 + |x|^{k}} $$
converges if and only if $|x| > 1$. I think that the best way to show the backwards direction is to assume we havve $|x| leq 1$ then maybe doing the integral test? But I didn't get anywhere with this. I don't relaly have an idea of how to do it with the other direction either.
Any help is appreciated.
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Dec 10 '18 at 2:10
user587192
1,775214
asked Dec 10 '18 at 1:50
joseph
4329
edited Dec 10 '18 at 2:10
user587192
1,775214
asked Dec 10 '18 at 1:50
joseph
4329
edited Dec 10 '18 at 2:10
user587192
1,775214
edited Dec 10 '18 at 2:10
user587192
1,775214
edited Dec 10 '18 at 2:10
user587192
1,775214
1,775214
asked Dec 10 '18 at 1:50
joseph
4329
asked Dec 10 '18 at 1:50
joseph
4329
asked Dec 10 '18 at 1:50
joseph
4329
4329
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Note that for $|x|>1$
$$
lim_{ktoinfty}frac{1+|x|^{k}}{1+|x|^{k+1}}
=lim_{ktoinfty}frac{|x|^{-k}+1}{|x|^{-k}+|x|}=frac{1}{|x|}<1.
$$
Now apply the ratio test.
For $|x|leqslant 1$,
$$
lim_{ktoinfty}frac{1}{1+|x|^{k}}
neq0.
$$
answered Dec 10 '18 at 1:56
user587192
1,775214
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that for $|x|>1$
$$
lim_{ktoinfty}frac{1+|x|^{k}}{1+|x|^{k+1}}
=lim_{ktoinfty}frac{|x|^{-k}+1}{|x|^{-k}+|x|}=frac{1}{|x|}<1.
$$
Now apply the ratio test.
For $|x|leqslant 1$,
$$
lim_{ktoinfty}frac{1}{1+|x|^{k}}
neq0.
$$
answered Dec 10 '18 at 1:56
user587192
1,775214
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
add a comment |
Note that for $|x|>1$
$$
lim_{ktoinfty}frac{1+|x|^{k}}{1+|x|^{k+1}}
=lim_{ktoinfty}frac{|x|^{-k}+1}{|x|^{-k}+|x|}=frac{1}{|x|}<1.
$$
Now apply the ratio test.
For $|x|leqslant 1$,
$$
lim_{ktoinfty}frac{1}{1+|x|^{k}}
neq0.
$$
answered Dec 10 '18 at 1:56
user587192
1,775214
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
add a comment |
Note that for $|x|>1$
$$
lim_{ktoinfty}frac{1+|x|^{k}}{1+|x|^{k+1}}
=lim_{ktoinfty}frac{|x|^{-k}+1}{|x|^{-k}+|x|}=frac{1}{|x|}<1.
$$
Now apply the ratio test.
For $|x|leqslant 1$,
$$
lim_{ktoinfty}frac{1}{1+|x|^{k}}
neq0.
$$
answered Dec 10 '18 at 1:56
user587192
1,775214
Note that for $|x|>1$
$$
lim_{ktoinfty}frac{1+|x|^{k}}{1+|x|^{k+1}}
=lim_{ktoinfty}frac{|x|^{-k}+1}{|x|^{-k}+|x|}=frac{1}{|x|}<1.
$$
Now apply the ratio test.
For $|x|leqslant 1$,
$$
lim_{ktoinfty}frac{1}{1+|x|^{k}}
neq0.
$$
answered Dec 10 '18 at 1:56
user587192
1,775214
edited Dec 10 '18 at 2:02
answered Dec 10 '18 at 1:56
user587192
1,775214
answered Dec 10 '18 at 1:56
user587192
1,775214
answered Dec 10 '18 at 1:56
user587192
1,775214
1,775214
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
add a comment |
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ?
– joseph
Dec 10 '18 at 6:42
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
@joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|leqslant1$, and show it diverges".
– user587192
Dec 10 '18 at 13:23
add a comment |
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