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Show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$
I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.
I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.
Can someone maybe help me please?
calculus real-analysis sequences-and-series
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
add a comment |
I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.
I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.
Can someone maybe help me please?
calculus real-analysis sequences-and-series
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31
add a comment |
I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.
I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.
Can someone maybe help me please?
calculus real-analysis sequences-and-series
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.
I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.
Can someone maybe help me please?
calculus real-analysis sequences-and-series
calculus real-analysis sequences-and-series
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
edited Nov 20 '18 at 18:39
Robert Z
93.3k1061132
93.3k1061132
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
asked Nov 20 '18 at 16:59
eager2learn
1,23211430
1,23211430
The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31
add a comment |
The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31
The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31
The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31
add a comment |
1 Answer
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Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
$$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
Can you take it from here?
P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
$$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
Can you take it from here?
P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
add a comment |
Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
$$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
Can you take it from here?
P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
add a comment |
Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
$$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
Can you take it from here?
P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
$$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
Can you take it from here?
P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
edited Nov 20 '18 at 17:11
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
answered Nov 20 '18 at 17:05
Robert Z
93.3k1061132
93.3k1061132
add a comment |
add a comment |
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The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31