Post History
#3: Post edited
by
Snoopy
·
2022-11-27T13:02:18Z (over 1 year ago)
> Let $\lfloor x floor$ be the maximum of the integer $n\le x$. Find the limit- $$
- \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
- ---
- I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here:
- - The limit is of the form $\infty\cdot 0$.
- - Integrating the power series for $\frac{1}{1-x}$ term by term, one has $\sum_{n=1}^\infty\frac{x^n}{n} = -\log(1-x)$ for small positive $x$ near $0$.
- - The expression can be rewritten as
- $$
- \lim_{x\to 0^+}e^{1/x}\left(-\log(1-x)-
- \sum^{\lfloor 1/x\rfloor}_{n=1}\frac{x^n}{n}
- \right)
- $$
- - Or as
- $$
- \lim_{x\to 0^+}\sum_{n=0}^\infty\frac{1}{n!x^n}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
- > Let $\lfloor x floor$ be the maximum integer $n\le x$. Find the limit
- $$
- \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
- ---
- I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here:
- - The limit is of the form $\infty\cdot 0$.
- - Integrating the power series for $\frac{1}{1-x}$ term by term, one has $\sum_{n=1}^\infty\frac{x^n}{n} = -\log(1-x)$ for small positive $x$ near $0$.
- - The expression can be rewritten as
- $$
- \lim_{x\to 0^+}e^{1/x}\left(-\log(1-x)-
- \sum^{\lfloor 1/x\rfloor}_{n=1}\frac{x^n}{n}
- \right)
- $$
- - Or as
- $$
- \lim_{x\to 0^+}\sum_{n=0}^\infty\frac{1}{n!x^n}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
#2: Post edited
by
Snoopy
·
2022-11-27T01:26:51Z (over 1 year ago)
Let $\lfloor x floor$ be the maximum of the integer $n\le x$. Find the limit- $$
- \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
- ---
- I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here:
- - The limit is of the form $\infty\cdot 0$.
- - Integrating the power series for $\frac{1}{1-x}$ term by term, one has $\sum_{n=1}^\infty\frac{x^n}{n} = -\log(1-x)$ for small positive $x$ near $0$.
- - The expression can be rewritten as
- $$
- \lim_{x\to 0^+}e^{1/x}\left(-\log(1-x)-
- \sum^{\lfloor 1/x\rfloor}_{n=1}\frac{x^n}{n}
- \right)
- $$
- - Or as
- $$
- \lim_{x\to 0^+}\sum_{n=0}^\infty\frac{1}{n!x^n}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
- > Let $\lfloor x floor$ be the maximum of the integer $n\le x$. Find the limit
- $$
- \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
- ---
- I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here:
- - The limit is of the form $\infty\cdot 0$.
- - Integrating the power series for $\frac{1}{1-x}$ term by term, one has $\sum_{n=1}^\infty\frac{x^n}{n} = -\log(1-x)$ for small positive $x$ near $0$.
- - The expression can be rewritten as
- $$
- \lim_{x\to 0^+}e^{1/x}\left(-\log(1-x)-
- \sum^{\lfloor 1/x\rfloor}_{n=1}\frac{x^n}{n}
- \right)
- $$
- - Or as
- $$
- \lim_{x\to 0^+}\sum_{n=0}^\infty\frac{1}{n!x^n}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
- $$
#1: Initial revision
by
Snoopy
·
2022-11-27T01:26:33Z (over 1 year ago)
Finding the limit $ \lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n} $
Let $\lfloor x \rfloor$ be the maximum of the integer $n\le x$. Find the limit
$$
\lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
$$
---
I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here:
- The limit is of the form $\infty\cdot 0$.
- Integrating the power series for $\frac{1}{1-x}$ term by term, one has $\sum_{n=1}^\infty\frac{x^n}{n} = -\log(1-x)$ for small positive $x$ near $0$.
- The expression can be rewritten as
$$
\lim_{x\to 0^+}e^{1/x}\left(-\log(1-x)-
\sum^{\lfloor 1/x\rfloor}_{n=1}\frac{x^n}{n}
\right)
$$
- Or as
$$
\lim_{x\to 0^+}\sum_{n=0}^\infty\frac{1}{n!x^n}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}
$$