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I've proved that

$$\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}$$

but I have no idea how to sum up this formula and take limit. Can anyone give me some tips? Thanks a lot!

UPD: Our instructor says that we can't use tools that are not taught in class before and I came up with another idea.

It seems that

$$\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}$$

and

$$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}) - (\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n})$$ and it is about $\ln(2n) - \ln(n)$ and thus the answer is $\ln(2)$. Am I consider it right?

Yunxuan Zhang
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Iloveolaf
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    Do you mean $\sum_{m=n+1}^{2n}$? And shouldn't the signs alternate? That is, $\frac1{n+1}-\frac1{n+2}+\frac1{n+3}-\cdots$. – ajotatxe Nov 19 '19 at 15:12
  • Here is my idea. Let $a = 1+\frac{1}{2}+...\frac{1}{n}$ and $b = 1+\frac{1}{2}+...\frac{1}{2n}$. Then $\sum_{m = 1}^{2n} \frac{(-1)^{m-1}}{m}$ will be b-a,and $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ will also be b-a. – Iloveolaf Nov 19 '19 at 15:20
  • The expression you wrote does not equal the sum. For one thing, the series is alternating in sign; for another the first term in the sum (with $m=1$) is $\tfrac11$, not $\tfrac{1}{n+1}$ – MPW Nov 19 '19 at 15:24
  • @MPW Excuse me, but I can't understand. Is there anything wrong with my proof? I can't see why the expression doesn't equal to the sum. – Iloveolaf Nov 19 '19 at 15:29
  • As written, the sum equals $$1-\tfrac12 + \tfrac13 - \tfrac14 + \cdots +\tfrac1{2n-1}-\tfrac1{2n},$$ not $$\tfrac{1}{n+1}+\tfrac{1}{n+2}+\tfrac{1}{n+3}+...+\tfrac{1}{2n}$$ Note the starting index you have in the sum is $m=1$. – MPW Nov 19 '19 at 15:37
  • @MPW But I can prove that the two formula is equal, and I prove it as above. So I'm wondering if there is anything wrong with my proof? – Iloveolaf Nov 19 '19 at 15:45
  • There's only so many ways I can say it. WHY DO YOU THINK THE FIRST TERM IS $\tfrac{1}{n+1}$???? AND DON'T YOU SEE THAT EVERY OTHER TERM IS NEGATIVE???? – MPW Nov 19 '19 at 15:57
  • @MPW Calm down. The two sums you wrote are actually equal. – Klaus Nov 19 '19 at 15:58
  • @Klaus : I'll be darned, so they are. My bad, three times. – MPW Nov 19 '19 at 16:08

2 Answers2

1

As mentioned in the comments, I would first rewrite the sum as

$$\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots+\frac{1}{2n-1}-\frac{1}{2n}\tag{1}$$

then recall the Taylor series for $\ln(x+1)$ $$\ln(x+1)=\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m}x^m=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\dots\tag{2}$$

If you let $n\to\infty$ in $(1)$, then you can match the alternating series in $(1)$ and $(2)$ by choosing a suitable value for $x$.

Axion004
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0

HINT Consider the Taylor series for the natural logarithm function around $x=1$.

gt6989b
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  • But our instructor said that we can't use methods not taught in class, we've only learned the basic concepts of integral so far. – Iloveolaf Nov 19 '19 at 15:46