How can I prove that: $$\frac{7}{12} < \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} < \frac{47}{60}$$
? I don't even know how to start solving this...
Asked
Active
Viewed 549 times
6
-
your sum has the value $\ln(2)$ – Dr. Sonnhard Graubner May 21 '16 at 15:12
-
2how did you compute that? – Shanon May 21 '16 at 15:13
-
This sum is also known as the "alternating harmonic series" which converges to $\ln(2)$ – Tesla May 21 '16 at 15:15
-
see also here http://www.efunda.com/math/exp_log/series_exp.cfm – Dr. Sonnhard Graubner May 21 '16 at 15:20
-
Possible duplicate of The sum $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)$ does not exist. – Mohan Sharma Jan 15 '18 at 08:51
4 Answers
7
First note that the series converges using Leibniz Test.
Next, denote by $S_N$ the partial sum $\sum_{n=1}^N\frac{(-1)^{n-1}}{n}$. Then, we must have $$S_{2N}<\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}<S_{2N+1}$$
Finally, we see that $\sum_{n=1}^4 \frac{(-1)^{n+1}}{n}=\frac{7}{12}$ and inasmuch as the next term is positive, the value of the series must exceed $7/12$. Similarly, we see that $\sum_{n=1}^5 \frac{(-1)^{n+1}}{n}=\frac{47}{60}$ and inasmuch as the next term is negative, the value of the series must be less than $47/60$.
Mark Viola
- 179,405
-
Is it also true that $S_{2N} < \sum_{n=1}^{k \in \mathbb{N}} \frac{(-1)^{n+1}}{n} < S_{2N+1}$ ? – Shanon May 21 '16 at 15:41
-
Shanon, I don't understand the meaning of the notation on the upper limit. Is $k$ any natural number? If so, then it cannot be true. – Mark Viola May 21 '16 at 15:57
-
You wrote that after $n=5$ the next term is negative. But the next term after that is positive so why does the value of the series must be less than $47/60$ ? – Shanon May 21 '16 at 16:22
-
-
-
3
If $s_{n}:=\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}}{k}$ then $\left(s_{2n-1}\right)_{n=1,2,\dots}$ is monotonically decreasing and $\left(s_{2n}\right)_{n=1,2,\dots}$ is monotonically increasing.
This with $s_{2n}<s_{2n-1}$.
So if you can find some $n$ with $\frac{7}{12}<s_{2n}<s_{2n-1}<\frac{47}{60}$ then you are ready.
drhab
- 151,093
-
I did not understand the last sentence... What do you mean I can findsome $n$ within that range? I understood that $S_{2n} < S_{2n-1}$ , can you elaborate more? – Shanon May 21 '16 at 15:32
-
$s_{2n+2}<s_{2n+1}<s_{2n-1}<\frac{47}{60}$ and so on. Likewise $s_{2n+1}>s_{2n+2}>s_{2n}>\frac7{12}$ and so on. So the inequality stays valid for larger $n$. – drhab May 21 '16 at 15:39
1
Hint your sum is integration of $\int (1+x)^{-1}$ after integration put $x=1$ . So series converges to $\ln(2)$
Archis Welankar
- 15,910
-
Since $x = 1$ lies at the boundary of the interval of convergence, there are some subtleties (involving Abel's Theorem) in giving a careful proof that the given series converges to $\log 2$. But this can certainly be done, so let's assume it: then how will you show that $\frac{7}{12} < \log 2 < \frac{47}{60}$? The easiest way to do this is to use the Alternating Series Test / Leibniz's Test as in the accepted answer. – Pete L. Clark May 21 '16 at 22:11
1
The manipulation $$\sum_{n\geq 0}\frac{(-1)^{n}}{n+1} = \int_{0}^{1}\sum_{n\geq 0}(-1)^n x^n\,dx = \int_{0}^{1}\frac{dx}{x+1} = \int_{0}^{1}\frac{dx}{2-x}=\sum_{n\geq 0}\frac{1}{(n+1)\,2^{n+1}}$$ gives a substantial convergence boost, and summation by parts gives a second boost: $$ \sum_{n\geq 0}\frac{1}{(n+1) 2^{n+1}} = 1-\sum_{n\geq 0}\frac{1}{(n+1)(n+2)2^{n+1}}\tag{1} $$ By evaluating $\sum_{n=0}^{4}\frac{1}{(n+1)2^{n+1}}$ and $1-\sum_{n=0}^{3}\frac{1}{(n+1)(n+2)2^{n+1}}$ we get the improved inequality:
$$ \color{red}{\frac{661}{960}}\leq \sum_{n\geq 0}\frac{(-1)^n}{n+1}\leq\color{red}{\frac{667}{960}}.\tag{2}$$
Jack D'Aurizio
- 353,855