I am confused on whether this leads to a finite value or not, but it seems the value of the series alternates around a certain interval, making it seem convergent. Please give me a valid proof as to if it's divergent or convergent.
Asked
Active
Viewed 111 times
0
-
Have you heard of the Alternating Series Test? – Dec 28 '16 at 00:27
-
No, please tell me. – Nafi04 Dec 28 '16 at 00:27
-
3The $N$-th partial sum is a Riemann sum for the integral $$\int_{0}^{1}\frac{dx}{1+x}$$ and since $f(x)=\frac{1}{1+x}$ is a continuous (hence integrable) function over $[0,1]$, your series is conditionally convergent to $$\log(2) = 0.6931471805599453\ldots $$ – Jack D'Aurizio Dec 28 '16 at 00:32
-
1See also: Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $. (Maybe also other questions linked there.) – Martin Sleziak Dec 26 '19 at 11:23
3 Answers
2
Notice that
$$\begin{align}\sum_{n=1}^\infty\frac{(-1)^{n+1}}n&=1-\frac12+\frac13-\frac14+\dots\\&=\left(1-\frac12\right)+\left(\frac13-\frac14\right)+\dots\\&>\frac12+0+0+0+\dots\\&=\frac12\\1-\frac12+\frac13-\frac14+\dots&=1+\left(-\frac12+\frac13\right)+\left(-\frac14+\frac15\right)+\dots\\&<1+0+0+0+\dots\\&=1\end{align}$$
Lastly, note that $\lim\limits_{n\to\infty}\frac{(-1)^{n+1}}n=0$, and that
$$\frac12<S_{2n}<S_{2n+1}<1,\quad S_{2n}-S_{2n+1}\to0\text{ as }n\to\infty$$
where $S_n$ is the $n$th partial sum.
Thus, it is convergent and
$$\frac12<\sum_{n=1}^\infty\frac{(-1)^{n+1}}n<1$$
The exact value can be found from the Taylor expansion of the natural logarithm:
$$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$$
Plugging in $x=1$ yields $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln(2)$
Simply Beautiful Art
- 74,685
-
@Nafi04 You are asking for the value of the series? It is $\ln(2)$, due to the Taylor expansion of $\ln(x)$. – Simply Beautiful Art Dec 28 '16 at 00:30
-
-
-
-
-
-
1
You can use the alternating series test. So consider the series: $\sum a_n$
if $a_n$ is: 1)decreasing, 2)positive 3)converges to $0$
Then by the Alternating Series Test, $\sum (-1)^n a_n $ converges
In your case $a_n$ is $\frac{1}{n}$ which satisfies the above conditions, then the series you originally had converges by the alternating series test.