Proof of Continuity on an Interval
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I'm having trouble proving that the function below is continuous on [1,e]
$$ sum_{n=1}^infty frac{1}{n3^n} ln(x) $$
Many answers I've seen already involve derivatives (which I cannot use) or just appear to have incomplete or incorrect answers.
Any help outlining a proof? I'm allowed to use the fact that the natural logarithm is continuous and increasing without proof.
real-analysis proof-writing
asked Dec 5 at 2:12
ktuggle
184
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I'm having trouble proving that the function below is continuous on [1,e]
$$ sum_{n=1}^infty frac{1}{n3^n} ln(x) $$
Many answers I've seen already involve derivatives (which I cannot use) or just appear to have incomplete or incorrect answers.
Any help outlining a proof? I'm allowed to use the fact that the natural logarithm is continuous and increasing without proof.
real-analysis proof-writing
asked Dec 5 at 2:12
ktuggle
184
1
You could argue that this series converges uniformly on $[1,e]$ (e.g. by Weierstrass M test or something else), so because the partial sums are continuous the uniform limit must also be continuous.
– Dave
Dec 5 at 2:23
3
Well. $sumfrac{1}{n3^n}ln x= text{const}cdot ln x$.
– Jacky Chong
Dec 5 at 2:32
Is it $x$ or $n$ ?
– Claude Leibovici
Dec 5 at 4:22
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having trouble proving that the function below is continuous on [1,e]
$$ sum_{n=1}^infty frac{1}{n3^n} ln(x) $$
Many answers I've seen already involve derivatives (which I cannot use) or just appear to have incomplete or incorrect answers.
Any help outlining a proof? I'm allowed to use the fact that the natural logarithm is continuous and increasing without proof.
real-analysis proof-writing
asked Dec 5 at 2:12
ktuggle
184
I'm having trouble proving that the function below is continuous on [1,e]
$$ sum_{n=1}^infty frac{1}{n3^n} ln(x) $$
Many answers I've seen already involve derivatives (which I cannot use) or just appear to have incomplete or incorrect answers.
Any help outlining a proof? I'm allowed to use the fact that the natural logarithm is continuous and increasing without proof.
real-analysis proof-writing
real-analysis proof-writing
asked Dec 5 at 2:12
ktuggle
184
asked Dec 5 at 2:12
ktuggle
184
asked Dec 5 at 2:12
ktuggle
184
asked Dec 5 at 2:12
ktuggle
184
asked Dec 5 at 2:12
ktuggle
184
184
1
You could argue that this series converges uniformly on $[1,e]$ (e.g. by Weierstrass M test or something else), so because the partial sums are continuous the uniform limit must also be continuous.
– Dave
Dec 5 at 2:23
3
Well. $sumfrac{1}{n3^n}ln x= text{const}cdot ln x$.
– Jacky Chong
Dec 5 at 2:32
Is it $x$ or $n$ ?
– Claude Leibovici
Dec 5 at 4:22
add a comment |
1
You could argue that this series converges uniformly on $[1,e]$ (e.g. by Weierstrass M test or something else), so because the partial sums are continuous the uniform limit must also be continuous.
– Dave
Dec 5 at 2:23
3
Well. $sumfrac{1}{n3^n}ln x= text{const}cdot ln x$.
– Jacky Chong
Dec 5 at 2:32
Is it $x$ or $n$ ?
– Claude Leibovici
Dec 5 at 4:22
1
1
You could argue that this series converges uniformly on $[1,e]$ (e.g. by Weierstrass M test or something else), so because the partial sums are continuous the uniform limit must also be continuous.
– Dave
Dec 5 at 2:23
You could argue that this series converges uniformly on $[1,e]$ (e.g. by Weierstrass M test or something else), so because the partial sums are continuous the uniform limit must also be continuous.
– Dave
Dec 5 at 2:23
3
3
Well. $sumfrac{1}{n3^n}ln x= text{const}cdot ln x$.
– Jacky Chong
Dec 5 at 2:32
Well. $sumfrac{1}{n3^n}ln x= text{const}cdot ln x$.
– Jacky Chong
Dec 5 at 2:32
Is it $x$ or $n$ ?
– Claude Leibovici
Dec 5 at 4:22
Is it $x$ or $n$ ?
– Claude Leibovici
Dec 5 at 4:22
add a comment |
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1
You could argue that this series converges uniformly on $[1,e]$ (e.g. by Weierstrass M test or something else), so because the partial sums are continuous the uniform limit must also be continuous.
– Dave
Dec 5 at 2:23
3
Well. $sumfrac{1}{n3^n}ln x= text{const}cdot ln x$.
– Jacky Chong
Dec 5 at 2:32
Is it $x$ or $n$ ?
– Claude Leibovici
Dec 5 at 4:22