$$\lim_{n\to\infty}\frac {1-\frac {1}{2} + \frac {1}{3} -\frac {1}{4}+ ... + \frac {1}{2n-1}-\frac{1}{2n}}{\frac {1}{n+1} + \frac {1}{n+2} + \frac {1}{n+3} + ... + \frac {1}{2n}}$$

I can express the value of the geometric sum of ${\frac {1}{2} + \frac {1}{4}+...+\frac {1}{2n}}$ but the others are ahead of me.

Putting both fraction parts under a common denominator makes that part tidy but the numerator seems to get way too complicated, which makes me think there is some simple way to do this.


2 Answers 2


Here's a solution with no calculus, just algebra. The numerator and the denominator are actually equal for all $n.$

Let $H_n=\sum_{k=1}^n \frac1{k}$ be the $k^{\text{th}}$ harmonic number.

Your denominator is $$D_n=\big(1+\frac12+\frac13+\dots+\frac1{2n}\big)-\big(1+\frac12+\frac13+\dots+\frac1{n}\big)=H_{2n}-H_n.$$

Your numerator is $$N_n=1-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}=\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}.$$


\begin{align}H_{2n}-N_n &=\big(1+\frac12+\frac13+\dots+\frac1{2n}\big)-\big(1-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}\big) \\ &= 2\cdot\!\!\!\!\!\sum_{\substack{1\le k \le 2n\\k\text{ is even}}}\frac1{k} \scriptsize\quad\quad{\text{ (because the odd terms cancel out and the even terms are doubled up)}} \\&=2\cdot\sum_{j=1}^n\frac1{2j} \\&=\sum_{j=1}^n \frac1{j} \\&=H_{n}, \end{align}

so $$ N_n=H_{2n}-H_n.$$

It follows that $N_n=D_n$ for all $n,$ so $N_n/D_n$ is a constant sequence with value always $1,$ and the limit is therefore $1.$

  • 3
    $\begingroup$ Very nice solution! $\endgroup$ Oct 10, 2016 at 5:27
  • 1
    $\begingroup$ Beautiful! You are right in that right now in class we are just beginning limits as a preparation for calculus. $\endgroup$
    – John Doe
    Oct 10, 2016 at 19:52
  • $\begingroup$ Just figured an even easier solution is $1-\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}=1+\frac{1}{2}+\frac{1}{3}+---+\frac{1}{2n}-2*(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})...=Denominator$ $\endgroup$
    – John Doe
    Oct 23, 2016 at 18:58

In the denominator we have:

$$\sum_{i=1}^{n} \frac{1}{n+i}$$

And as $n \to \infty$ this is:

$$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n+i}=\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{1+(0+i \frac{1-0}{n})}\frac{1-0}{n}=\int_{0}^{1} \frac{1}{1+x} dx=\ln (1+1)-\ln(1+0)=\ln (2)$$

For the numerator as $n \to \infty$, considering the geometric series $\sum_{n=0}^{\infty} u^n=\frac{1}{1-u}$ for $|u|<1$, substituting in $-x$ for $u$ we have:

$$\frac{1}{1-(-x)}=\sum_{n=0}^{\infty} (-1)^nx^n$$

Then integrate both sides to get

$$\ln(1+x)=C+\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n+1}}{n+1}=C+\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$

Let $x=0 \implies C=0$, so:

$$\ln (1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$

Because the series converges for $x=1$ by the alternating series test, we are justified in letting $x \to 1^-$ to get:

$$\ln (2)=\sum_{n=1}^{\infty} (-1)^n\frac{1}{n}$$

Hence the limit is:

$$\frac{\ln 2}{\ln 2}=1$$

  • 1
    $\begingroup$ Ah, sorry, but this goes way over my head. There has to exist a solution without integrals or calculus as at this point it isn't assumed we know these. $\endgroup$
    – John Doe
    Oct 10, 2016 at 4:20
  • $\begingroup$ @JohnDoe If you thought of the problem yourself, then realize many problems can be made with the tools you have, though many cannot be solved with those same tools. If it was a problem given to you by, say, your teacher, then you are taking a very strange class IMO. $\endgroup$ Oct 10, 2016 at 12:26
  • $\begingroup$ @SimpleArt See my answer for an elementary solution. Somewhat surprisingly, it's actually a good problem for a class just starting limits, or for a pre-calculus class that's doing sums and combinatorics. $\endgroup$ Oct 10, 2016 at 17:05

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