Proving a limit exists under some conditions
$begingroup$
Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
infty)$. Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
ldots$. Prove that
$$lim_{ntoinfty} F_{n} $$
exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.
Hint: Consider $F(n + 1) - F(n)$
I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.
real-analysis calculus sequences-and-series limits analysis
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
$endgroup$
add a comment |
$begingroup$
Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
infty)$. Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
ldots$. Prove that
$$lim_{ntoinfty} F_{n} $$
exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.
Hint: Consider $F(n + 1) - F(n)$
I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.
real-analysis calculus sequences-and-series limits analysis
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
$endgroup$
add a comment |
$begingroup$
Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
infty)$. Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
ldots$. Prove that
$$lim_{ntoinfty} F_{n} $$
exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.
Hint: Consider $F(n + 1) - F(n)$
I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.
real-analysis calculus sequences-and-series limits analysis
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
$endgroup$
Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
infty)$. Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
ldots$. Prove that
$$lim_{ntoinfty} F_{n} $$
exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.
Hint: Consider $F(n + 1) - F(n)$
I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.
real-analysis calculus sequences-and-series limits analysis
real-analysis calculus sequences-and-series limits analysis
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
edited Dec 14 '18 at 18:04
Dillain Smith
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
asked Dec 14 '18 at 17:54
Dillain SmithDillain Smith
496
496
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint
Note that
$$
sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
$$
where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
hint
For $nge 0,$
$$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$
$$(forall tin[n,n+1]);$$
$$f(n+1)le f(t)le f(n)$$
and
$$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)
If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$
Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.
From this we can see that the following comparison is true
$$
F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
$$
now we can take the limit of both sides and see that
$$
lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
$$
Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.
To prove the other direction the method is the same, except we say that
$$
sum_{k=1}^n f(k) < int_0^n f(t)dt
$$
since it is not a "right-Riemann Sum"
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Note that
$$
sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
$$
where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Hint
Note that
$$
sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
$$
where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Hint
Note that
$$
sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
$$
where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
Hint
Note that
$$
sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
$$
where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
answered Dec 14 '18 at 18:02
Foobaz JohnFoobaz John
21.6k41352
21.6k41352
add a comment |
add a comment |
$begingroup$
hint
For $nge 0,$
$$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$
$$(forall tin[n,n+1]);$$
$$f(n+1)le f(t)le f(n)$$
and
$$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
For $nge 0,$
$$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$
$$(forall tin[n,n+1]);$$
$$f(n+1)le f(t)le f(n)$$
and
$$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
For $nge 0,$
$$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$
$$(forall tin[n,n+1]);$$
$$f(n+1)le f(t)le f(n)$$
and
$$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hint
For $nge 0,$
$$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$
$$(forall tin[n,n+1]);$$
$$f(n+1)le f(t)le f(n)$$
and
$$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 14 '18 at 18:25
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
$begingroup$
I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)
If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$
Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.
From this we can see that the following comparison is true
$$
F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
$$
now we can take the limit of both sides and see that
$$
lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
$$
Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.
To prove the other direction the method is the same, except we say that
$$
sum_{k=1}^n f(k) < int_0^n f(t)dt
$$
since it is not a "right-Riemann Sum"
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
$endgroup$
add a comment |
$begingroup$
I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)
If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$
Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.
From this we can see that the following comparison is true
$$
F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
$$
now we can take the limit of both sides and see that
$$
lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
$$
Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.
To prove the other direction the method is the same, except we say that
$$
sum_{k=1}^n f(k) < int_0^n f(t)dt
$$
since it is not a "right-Riemann Sum"
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
$endgroup$
add a comment |
$begingroup$
I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)
If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$
Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.
From this we can see that the following comparison is true
$$
F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
$$
now we can take the limit of both sides and see that
$$
lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
$$
Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.
To prove the other direction the method is the same, except we say that
$$
sum_{k=1}^n f(k) < int_0^n f(t)dt
$$
since it is not a "right-Riemann Sum"
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
$endgroup$
I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)
If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$
Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.
From this we can see that the following comparison is true
$$
F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
$$
now we can take the limit of both sides and see that
$$
lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
$$
Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.
To prove the other direction the method is the same, except we say that
$$
sum_{k=1}^n f(k) < int_0^n f(t)dt
$$
since it is not a "right-Riemann Sum"
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
answered Dec 15 '18 at 0:33
wjmccannwjmccann
654118
654118
add a comment |
add a comment |
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