Mixing
Separable Hilbert Space
$begingroup$
Let $mathcal{H}$ be a separable Hilbert space with complete orthonormal system $e_{1},e_{2},ldots$
(a) Let $T:mathcal{H}rightarrowmathcal{H}$ be a continuous linear operator. Suppose that there is a sequence $(v_{i})_{igeq1}$ of strictly positive numbers, and numbers $lambda,mugeq 0$, with the property that,
begin{align}
sum_{j=1}^{infty} |(Te_{j}|e_{i})|v_{j} &leqlambda v_{i}\[3pt]
sum_{i=1}^{infty} |(Te_{j}|e_{i})|v_{i} &leqmu v_{j},
end{align}
for all $i,jgeq 1$. Show that $|T|leqsqrt{lambdamu}$.
Hint: Apply Parseval's Identity to $|Tx|^{2}$. Expand $x$ into it's Fourier series to see that
begin{align}
|(Tx|e_{i})|leqsum_{j=1}^{infty}|(x|e_{j})||(Te_{j}|e_{i})|.
end{align}
Then write $|(x|e_{j})||(Te_{j}|e_{i})|=bigg(|(x|e_{j})|v_{j}^{-1/2}|(Te_{j}|e_{i})|^{1/2}bigg)bigg(|(Te_{j}|e_{i})|^{1/2}v_{j}^{1/2}bigg)$.
(b) Suppose that $T:mathcal{H}rightarrowmathcal{H}$ is a continuous linear operator with,
begin{align}
(Te_{j}|e_{i})=(i+j-1)^{-1}.
end{align}
Show that $|T|leqpi$.
Hint: Use (a) with $v_{j}=(j-frac{1}{2})^{-1/2}$, and approximate sums by integrals.
I have completed part (a), but I am confused by part (b) of the question. When the hint mentions to approximate sums by integrals, is it talking about the inequalities given in part (a)? or part of the working for part (a) where the inequality in the hint comes in to play?
functional-analysis inequality inner-product-space orthonormal
edited Jan 28 at 22:05
Yaddle
3,136829
asked Jan 28 at 4:10
JackJack
937
$endgroup$
add a comment |
$begingroup$
Let $mathcal{H}$ be a separable Hilbert space with complete orthonormal system $e_{1},e_{2},ldots$
(a) Let $T:mathcal{H}rightarrowmathcal{H}$ be a continuous linear operator. Suppose that there is a sequence $(v_{i})_{igeq1}$ of strictly positive numbers, and numbers $lambda,mugeq 0$, with the property that,
begin{align}
sum_{j=1}^{infty} |(Te_{j}|e_{i})|v_{j} &leqlambda v_{i}\[3pt]
sum_{i=1}^{infty} |(Te_{j}|e_{i})|v_{i} &leqmu v_{j},
end{align}
for all $i,jgeq 1$. Show that $|T|leqsqrt{lambdamu}$.
Hint: Apply Parseval's Identity to $|Tx|^{2}$. Expand $x$ into it's Fourier series to see that
begin{align}
|(Tx|e_{i})|leqsum_{j=1}^{infty}|(x|e_{j})||(Te_{j}|e_{i})|.
end{align}
Then write $|(x|e_{j})||(Te_{j}|e_{i})|=bigg(|(x|e_{j})|v_{j}^{-1/2}|(Te_{j}|e_{i})|^{1/2}bigg)bigg(|(Te_{j}|e_{i})|^{1/2}v_{j}^{1/2}bigg)$.
(b) Suppose that $T:mathcal{H}rightarrowmathcal{H}$ is a continuous linear operator with,
begin{align}
(Te_{j}|e_{i})=(i+j-1)^{-1}.
end{align}
Show that $|T|leqpi$.
Hint: Use (a) with $v_{j}=(j-frac{1}{2})^{-1/2}$, and approximate sums by integrals.
I have completed part (a), but I am confused by part (b) of the question. When the hint mentions to approximate sums by integrals, is it talking about the inequalities given in part (a)? or part of the working for part (a) where the inequality in the hint comes in to play?
functional-analysis inequality inner-product-space orthonormal
edited Jan 28 at 22:05
Yaddle
3,136829
asked Jan 28 at 4:10
JackJack
937
$endgroup$
add a comment |
$begingroup$
Let $mathcal{H}$ be a separable Hilbert space with complete orthonormal system $e_{1},e_{2},ldots$
(a) Let $T:mathcal{H}rightarrowmathcal{H}$ be a continuous linear operator. Suppose that there is a sequence $(v_{i})_{igeq1}$ of strictly positive numbers, and numbers $lambda,mugeq 0$, with the property that,
begin{align}
sum_{j=1}^{infty} |(Te_{j}|e_{i})|v_{j} &leqlambda v_{i}\[3pt]
sum_{i=1}^{infty} |(Te_{j}|e_{i})|v_{i} &leqmu v_{j},
end{align}
for all $i,jgeq 1$. Show that $|T|leqsqrt{lambdamu}$.
Hint: Apply Parseval's Identity to $|Tx|^{2}$. Expand $x$ into it's Fourier series to see that
begin{align}
|(Tx|e_{i})|leqsum_{j=1}^{infty}|(x|e_{j})||(Te_{j}|e_{i})|.
end{align}
Then write $|(x|e_{j})||(Te_{j}|e_{i})|=bigg(|(x|e_{j})|v_{j}^{-1/2}|(Te_{j}|e_{i})|^{1/2}bigg)bigg(|(Te_{j}|e_{i})|^{1/2}v_{j}^{1/2}bigg)$.
(b) Suppose that $T:mathcal{H}rightarrowmathcal{H}$ is a continuous linear operator with,
begin{align}
(Te_{j}|e_{i})=(i+j-1)^{-1}.
end{align}
Show that $|T|leqpi$.
Hint: Use (a) with $v_{j}=(j-frac{1}{2})^{-1/2}$, and approximate sums by integrals.
I have completed part (a), but I am confused by part (b) of the question. When the hint mentions to approximate sums by integrals, is it talking about the inequalities given in part (a)? or part of the working for part (a) where the inequality in the hint comes in to play?
functional-analysis inequality inner-product-space orthonormal
edited Jan 28 at 22:05
Yaddle
3,136829
asked Jan 28 at 4:10
JackJack
937
$endgroup$
Let $mathcal{H}$ be a separable Hilbert space with complete orthonormal system $e_{1},e_{2},ldots$
(a) Let $T:mathcal{H}rightarrowmathcal{H}$ be a continuous linear operator. Suppose that there is a sequence $(v_{i})_{igeq1}$ of strictly positive numbers, and numbers $lambda,mugeq 0$, with the property that,
begin{align}
sum_{j=1}^{infty} |(Te_{j}|e_{i})|v_{j} &leqlambda v_{i}\[3pt]
sum_{i=1}^{infty} |(Te_{j}|e_{i})|v_{i} &leqmu v_{j},
end{align}
for all $i,jgeq 1$. Show that $|T|leqsqrt{lambdamu}$.
Hint: Apply Parseval's Identity to $|Tx|^{2}$. Expand $x$ into it's Fourier series to see that
begin{align}
|(Tx|e_{i})|leqsum_{j=1}^{infty}|(x|e_{j})||(Te_{j}|e_{i})|.
end{align}
Then write $|(x|e_{j})||(Te_{j}|e_{i})|=bigg(|(x|e_{j})|v_{j}^{-1/2}|(Te_{j}|e_{i})|^{1/2}bigg)bigg(|(Te_{j}|e_{i})|^{1/2}v_{j}^{1/2}bigg)$.
(b) Suppose that $T:mathcal{H}rightarrowmathcal{H}$ is a continuous linear operator with,
begin{align}
(Te_{j}|e_{i})=(i+j-1)^{-1}.
end{align}
Show that $|T|leqpi$.
Hint: Use (a) with $v_{j}=(j-frac{1}{2})^{-1/2}$, and approximate sums by integrals.
I have completed part (a), but I am confused by part (b) of the question. When the hint mentions to approximate sums by integrals, is it talking about the inequalities given in part (a)? or part of the working for part (a) where the inequality in the hint comes in to play?
functional-analysis inequality inner-product-space orthonormal
functional-analysis inequality inner-product-space orthonormal
edited Jan 28 at 22:05
Yaddle
3,136829
asked Jan 28 at 4:10
JackJack
937
edited Jan 28 at 22:05
Yaddle
3,136829
asked Jan 28 at 4:10
JackJack
937
edited Jan 28 at 22:05
Yaddle
3,136829
edited Jan 28 at 22:05
Yaddle
3,136829
edited Jan 28 at 22:05
Yaddle
3,136829
3,136829
asked Jan 28 at 4:10
JackJack
937
asked Jan 28 at 4:10
JackJack
937
asked Jan 28 at 4:10
JackJack
937
937
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Starting from the hint one obtains
$$ sum_{j = 1}^infty vert (Te_j, e_i) vert v_j = sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}}.$$
Now the summands behave like $a_j := ((i + j - 1) sqrt{j - 1/2})^{-1} = mathcal O(j^{-3/2})$ for each $i in mathbb N$ , so the sum converges for all $i in mathbb N$. Moreover, the sequence $(a_j)_{j in mathbb N}$ is decreasing for each $i in mathbb N$. So by the integral test the integral
$$ int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx$$ is finite for all $i in mathbb N$ and one has
begin{align*}
sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} &leq frac{sqrt 2}{i} + int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx \
&= frac{sqrt 2}{i} + 2 frac{arctan(sqrt{2 i - 1})}{sqrt{i - 1/2}} \
&= left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i.
end{align*}
Now the first factor in the estimate still depends on $i$. But the function
$$ g : x mapsto frac{sqrt 2 sqrt{x - 1/2}}{x} + 2 arctan(sqrt{2 x - 1})$$
is increasing on $[1/2, infty)$ and one clearly has that $lim_{x to infty} g(x) = pi$. So finally we obtain the estimate
begin{align*}
sum_{j = 1}^infty vert (Te_j, e_i) vert v_j &= sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} \
&leq left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i leq pi v_i.
end{align*}
By interchanging the roles of $i$ and $j$ one obtains analogously that
$$sum_{i = 1}^infty vert (Te_j, e_i) vert v_j leq pi v_j. $$
So (a) yields that $Vert T Vert leq sqrt{pi^2} = pi$ which is the desired estimate. I hope it got clear :)
answered Jan 28 at 23:03
YaddleYaddle
3,136829
$endgroup$
1
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Starting from the hint one obtains
$$ sum_{j = 1}^infty vert (Te_j, e_i) vert v_j = sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}}.$$
Now the summands behave like $a_j := ((i + j - 1) sqrt{j - 1/2})^{-1} = mathcal O(j^{-3/2})$ for each $i in mathbb N$ , so the sum converges for all $i in mathbb N$. Moreover, the sequence $(a_j)_{j in mathbb N}$ is decreasing for each $i in mathbb N$. So by the integral test the integral
$$ int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx$$ is finite for all $i in mathbb N$ and one has
begin{align*}
sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} &leq frac{sqrt 2}{i} + int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx \
&= frac{sqrt 2}{i} + 2 frac{arctan(sqrt{2 i - 1})}{sqrt{i - 1/2}} \
&= left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i.
end{align*}
Now the first factor in the estimate still depends on $i$. But the function
$$ g : x mapsto frac{sqrt 2 sqrt{x - 1/2}}{x} + 2 arctan(sqrt{2 x - 1})$$
is increasing on $[1/2, infty)$ and one clearly has that $lim_{x to infty} g(x) = pi$. So finally we obtain the estimate
begin{align*}
sum_{j = 1}^infty vert (Te_j, e_i) vert v_j &= sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} \
&leq left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i leq pi v_i.
end{align*}
By interchanging the roles of $i$ and $j$ one obtains analogously that
$$sum_{i = 1}^infty vert (Te_j, e_i) vert v_j leq pi v_j. $$
So (a) yields that $Vert T Vert leq sqrt{pi^2} = pi$ which is the desired estimate. I hope it got clear :)
answered Jan 28 at 23:03
YaddleYaddle
3,136829
$endgroup$
1
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
add a comment |
$begingroup$
Starting from the hint one obtains
$$ sum_{j = 1}^infty vert (Te_j, e_i) vert v_j = sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}}.$$
Now the summands behave like $a_j := ((i + j - 1) sqrt{j - 1/2})^{-1} = mathcal O(j^{-3/2})$ for each $i in mathbb N$ , so the sum converges for all $i in mathbb N$. Moreover, the sequence $(a_j)_{j in mathbb N}$ is decreasing for each $i in mathbb N$. So by the integral test the integral
$$ int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx$$ is finite for all $i in mathbb N$ and one has
begin{align*}
sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} &leq frac{sqrt 2}{i} + int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx \
&= frac{sqrt 2}{i} + 2 frac{arctan(sqrt{2 i - 1})}{sqrt{i - 1/2}} \
&= left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i.
end{align*}
Now the first factor in the estimate still depends on $i$. But the function
$$ g : x mapsto frac{sqrt 2 sqrt{x - 1/2}}{x} + 2 arctan(sqrt{2 x - 1})$$
is increasing on $[1/2, infty)$ and one clearly has that $lim_{x to infty} g(x) = pi$. So finally we obtain the estimate
begin{align*}
sum_{j = 1}^infty vert (Te_j, e_i) vert v_j &= sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} \
&leq left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i leq pi v_i.
end{align*}
By interchanging the roles of $i$ and $j$ one obtains analogously that
$$sum_{i = 1}^infty vert (Te_j, e_i) vert v_j leq pi v_j. $$
So (a) yields that $Vert T Vert leq sqrt{pi^2} = pi$ which is the desired estimate. I hope it got clear :)
answered Jan 28 at 23:03
YaddleYaddle
3,136829
$endgroup$
1
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
add a comment |
$begingroup$
Starting from the hint one obtains
$$ sum_{j = 1}^infty vert (Te_j, e_i) vert v_j = sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}}.$$
Now the summands behave like $a_j := ((i + j - 1) sqrt{j - 1/2})^{-1} = mathcal O(j^{-3/2})$ for each $i in mathbb N$ , so the sum converges for all $i in mathbb N$. Moreover, the sequence $(a_j)_{j in mathbb N}$ is decreasing for each $i in mathbb N$. So by the integral test the integral
$$ int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx$$ is finite for all $i in mathbb N$ and one has
begin{align*}
sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} &leq frac{sqrt 2}{i} + int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx \
&= frac{sqrt 2}{i} + 2 frac{arctan(sqrt{2 i - 1})}{sqrt{i - 1/2}} \
&= left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i.
end{align*}
Now the first factor in the estimate still depends on $i$. But the function
$$ g : x mapsto frac{sqrt 2 sqrt{x - 1/2}}{x} + 2 arctan(sqrt{2 x - 1})$$
is increasing on $[1/2, infty)$ and one clearly has that $lim_{x to infty} g(x) = pi$. So finally we obtain the estimate
begin{align*}
sum_{j = 1}^infty vert (Te_j, e_i) vert v_j &= sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} \
&leq left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i leq pi v_i.
end{align*}
By interchanging the roles of $i$ and $j$ one obtains analogously that
$$sum_{i = 1}^infty vert (Te_j, e_i) vert v_j leq pi v_j. $$
So (a) yields that $Vert T Vert leq sqrt{pi^2} = pi$ which is the desired estimate. I hope it got clear :)
answered Jan 28 at 23:03
YaddleYaddle
3,136829
$endgroup$
Starting from the hint one obtains
$$ sum_{j = 1}^infty vert (Te_j, e_i) vert v_j = sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}}.$$
Now the summands behave like $a_j := ((i + j - 1) sqrt{j - 1/2})^{-1} = mathcal O(j^{-3/2})$ for each $i in mathbb N$ , so the sum converges for all $i in mathbb N$. Moreover, the sequence $(a_j)_{j in mathbb N}$ is decreasing for each $i in mathbb N$. So by the integral test the integral
$$ int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx$$ is finite for all $i in mathbb N$ and one has
begin{align*}
sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} &leq frac{sqrt 2}{i} + int_1^infty frac{1}{(i + x - 1) sqrt{x - 1/2}} , dx \
&= frac{sqrt 2}{i} + 2 frac{arctan(sqrt{2 i - 1})}{sqrt{i - 1/2}} \
&= left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i.
end{align*}
Now the first factor in the estimate still depends on $i$. But the function
$$ g : x mapsto frac{sqrt 2 sqrt{x - 1/2}}{x} + 2 arctan(sqrt{2 x - 1})$$
is increasing on $[1/2, infty)$ and one clearly has that $lim_{x to infty} g(x) = pi$. So finally we obtain the estimate
begin{align*}
sum_{j = 1}^infty vert (Te_j, e_i) vert v_j &= sum_{j = 1}^infty frac{1}{(i + j - 1) sqrt{j - 1/2}} \
&leq left(frac{sqrt 2 sqrt{i - 1/2}}{i} + 2 arctan(sqrt{2 i - 1}) right) v_i leq pi v_i.
end{align*}
By interchanging the roles of $i$ and $j$ one obtains analogously that
$$sum_{i = 1}^infty vert (Te_j, e_i) vert v_j leq pi v_j. $$
So (a) yields that $Vert T Vert leq sqrt{pi^2} = pi$ which is the desired estimate. I hope it got clear :)
answered Jan 28 at 23:03
YaddleYaddle
3,136829
answered Jan 28 at 23:03
YaddleYaddle
3,136829
answered Jan 28 at 23:03
YaddleYaddle
3,136829
answered Jan 28 at 23:03
YaddleYaddle
3,136829
3,136829
1
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
add a comment |
1
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
1
1
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
This is fantastic. Can you please clarify where the $frac{sqrt{2}}{i}$ term in the integral approximation of the sum comes from? I am not that familiar with approximating sums by integrals.
$endgroup$
– Jack
Jan 29 at 1:18
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
@Jack Look at the first remark in the article on the integral test. From there you obtain the estimate $sum_{j = 1}^infty g(j) leq g(1) + int_1^infty g(x) , dx$. The term $sqrt 2 / i$ is just $g(1)$.
$endgroup$
– Yaddle
Jan 29 at 8:30
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
$begingroup$
Thank you. Got it now.
$endgroup$
– Jack
Jan 30 at 1:39
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