Investigate the convergence or divergence of the series

$$\sum_{n=1}^\infty (-1)^{n-1}n=1-2+3-4 + \cdots$$

Could anyone help me with this problem?

  • 2
    $\begingroup$ Look at the partial sums: $1,-1,2,-2,3,-3,\dots$. Do they converge to anything? $\endgroup$ – Brian M. Scott May 21 '12 at 22:33
  • $\begingroup$ Do the terms you're adding converge to zero? $\endgroup$ – David Mitra May 21 '12 at 22:49
  • $\begingroup$ It is Abel summable to 1/4. Also, it is the series obtained as the Cauchy product of $1-1+1-1+1-1+\cdots$ with itself, and the latter series is Abel summable to 1/2. $\endgroup$ – Jonas Meyer May 21 '12 at 22:56

Hint: A necessary condition for series $\sum_{n=1}^\infty a_n$ to converge is $\lim_{n\to\infty} a_n=0$. As you clearly see, $\lim_{n\to\infty} (-1)^{n-1}n$ doesn't exist, in particular it is not $0$.

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  • $\begingroup$ What do you want to know is where it diverges? $\endgroup$ – Daniela del Carmen May 21 '12 at 23:00
  • $\begingroup$ @Daniela del Carmen: What do you mean by "where it diverges"? It is a number series, so either it diverges or converges, there is no region of convergence/divergence. $\endgroup$ – Dennis Gulko May 22 '12 at 0:05

HINT Look at the partial sum $\displaystyle S_m = \sum_{k=1}^{m} (-1)^{k-1}k$. See what happens when $m$ is odd and even.

If you are interested in more details, move your mouse over the gray area below.

The sum does not converge in the usual sense. By usual sense, we mean that if we interpret the sum as the limit of the sequence of partial sums then the sum does not converge. This is so since $$S_m = \sum_{k=1}^{m} (-1)^{k-1}k = \begin{cases} \frac{m+1}{2} & \text{ when $m$ is odd}\\ -\frac{m}{2} & \text{ when $m$ is even}\end{cases}$$ and as $m$ becomes large it oscillates between large positive value and negative values depending on the parity of $m$. Hence, $\lim_{n \rightarrow \infty} S_n$ does not exist.

Move your mouse over the gray area below for other interpretations of this infinite sum.

However, there are other ways to interpret this sum. Another popular way of defining convergence is to look at the Cesàro sum. The Cesàro sum is defined as the limit of the average of the sequence of partial sums i.e. if we let $$\tilde{S}_n = \dfrac{\displaystyle\sum_{m=1}^{n} S_m}{n},$$ then we have that $$\tilde{S}_n = \begin{cases} \frac{n+1}{2n} & \text{if $n$ is odd},\\ 0& \text{ if $n$ is even},\end{cases}$$ which implies that the series again doesn't converge. You can however smoothen it further by looking at $$\hat{S}_{n} = \dfrac{\displaystyle \sum_{m=1}^{n} \tilde{S}_m}{n}$$ Now you can show that $\hat{S}_n \rightarrow \frac14$. This sum is termed as Holder sum. If this limit were again to be infinite, then you further smooth it by taking its average. These are also termed Holder sums and denoted as $H(k)$ where $k$ is the number of times we do the smoothing. For instance, the usual sum is $H(0)$, the Cesaro sum is $H(1)$, the sum we did to get the limit as $\frac14$ is $H(2)$ and so on.

Analytic continuation of power series.

Another interpretation is through analytic continuation of the function $$f(x) = 1-2x+3x^2 - 4x^3 + \cdots, \, \lvert x \rvert < 1.$$The function converges to $\dfrac1{(1+x)^2}$ on the interval $(-1,1)$. Not surprisingly, the analytic continuation of the function $f(x)$ on the entire complex plane is given by $f_{\text{ac}}(x) = \dfrac1{(1+x)^2}$, where $x \in \mathbb{C}\backslash \{-1\}$. Hence, if we plugin the value of $x=1$ in $f_{\text{ac}}(x)$, we get the value of $\frac14$. Hence, $$1 - 2 + 3 - 4 + \cdots \underset{\text{ac}}{=} \frac14.$$

Zeta regularization:

The $\zeta$ regularization technique, which is another popular technique to regularize such infinite summations, also based on analytic continuation, can also be used here to get the value of $1-2+3-4+\cdots$. Consider $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots$$ and $$f(s) = 1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots$$The series $g(s)$ converges for Real$(s) > 0$ and converges absolutely for Real$(s) >1$. The series $f(s)$ converges for Real$(s) > 1$. In the region, Real$(s)>1$, we have that $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots\\ = \left(1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots \right) - \left(\frac2{2^s} + \frac2{4^s} + \frac2{6^s} + \frac2{8^s} + \cdots \right)\\ = f(s) - \frac1{2^{s-1}} f(s) = \left(1 - 2^{1-s} \right)f(s).$$ Now analytic continuation of $g(s)$ gives us $\eta(s)$ such that $\eta(s) = (1-2^{1-s}) \zeta(s)$, where $\zeta(s)$ is the analytic continuation of $f(s)$. Plugging in $s=-1$, we get that $$1 - 2 + 3 - 4 +\cdots \underset{\text{ac}}{=} \eta(-1) = (1-2^{1-(-1)}) \zeta(-1) = (1-4) \times \frac{-1}{12} = \frac14.$$ There are also other closely related regularization techniques like Borel summation, Ramanujan summation which assign a finite value to sums which do not converge. All these different techniques assign a value of $\dfrac14$ for this series.

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You could use the limit test for convergence, but you don't actually need that here at all. The first few partial sums are $1, -1, 2, -2, \ldots$. See if you can prove the pattern continues.

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