I just came across a way that the amazing mathematician Ramanujan developed of assigning a value to certain divergent series. I found it interesting, so I want to share a short summary of it here. It is based on the Euler–Maclaurin formula
\begin{align*} \sum_{k=\alpha}^\beta f(k) &= \int_\alpha^\beta f(t)\,dt + \frac{f(\alpha)+f(\beta)}{2} \\ &\quad+ \sum_{k=1}^n \frac{B_{2k}}{(2k)!}\Bigl(f^{(2k-1)}(\beta)-f^{(2k-1)}(\alpha)\Bigr) + R_n \end{align*}where $B_{2k}$ denotes the $(2k)$th Bernoulli number, $f$ has $2n+1$ continuous derivatives on $[\alpha,\beta]$ with $\alpha$, $\beta$, and $n\geq0$ being integers, and $R_n$ is the remainder term given by
\[ R_n = \int_\alpha^\beta \frac{B_{2n+1}(t-\lfloor t\rfloor)}{(2n+1)!}f^{(2n+1)}(t)\,dt . \] Continue reading →

Shortly after my last post I realized there was a simpler way of determining the exact value of the series $\sum_{n=1}^\infty\cos n/n$. Instead of following the method I previously described which required the intricate analysis of some integrals, one can simply use the formula

\[ \sum_{n=1}^\infty\frac{a^n}{n} = -\ln(1-a) \]

which is valid for $a\in\mathbb{C}$ which satisfies $\lvert a\rvert\leq 1$ and $a\neq 1$. This comes from a simple rewriting of the so-called Mercator series (replace $x$ with $-x$ in the Taylor series of $\ln(1+x)$ and then take the negative).

Then we have

\begin{align*}
\sum_{n=1}^\infty\frac{\cos n}{n} &= \sum_{n=1}^\infty\frac{e^{in}+e^{-in}}{2n} \\
&= -\bigl(\ln(1-e^i)+\ln(1-e^{-i})\bigr)/2 \\
&= -\ln\bigl((1-e^i)(1-e^{-i})\bigr)/2 \\
&= -\ln(2-e^i-e^{-i})/2 \\
&= -\ln(2-2\cos1)/2 \\
&\approx 0.0420195
\end{align*}

since $\lvert e^i\rvert=\lvert e^{-i}\rvert=1$, but $e^i\neq1$ and $e^{-i}\neq1$.

I’ve previously discussed various aspects of the “trigonometric harmonic series” $\sum_{n=1}^\infty\cos n/n$, and in particular showed that the series is conditionally convergent. However, we haven’t found the actual value it converges to; our argument only shows that the value must be smaller than about $2.54$ in absolute value. In this post, I’ll give a closed-form expression for the exact value that the series converges to.

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The purpose of this post is to show the interesting identity

\[ \sin(x)^2-\sin(y)^2 = \sin(x+y)\sin(x-y) . \]

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In this post I want to prove a lemma which gives a closed-form expression for the summation $\sum_{n=1}^m\sin(nx)$. The method of proof has come up before; it uses basic algebra, the complex exponential expression for sine, and the summation of a geometric series formula.

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We’ve discussed before the “trigonometric harmonic” series $\sum_{n=1}^\infty\cos n/n$. In particular, we showed that the series converges (conditionally). The argument involved the partial sums of the sequence $\{\cos n\}_{n=1}^\infty$, and we denoted these by $C(m)$. The closed-form expression we found for $C(m)$ involved the quantity $\cos m-\cos(m+1)$; in this post we show that this expression can also be written in the alternative form $2\sin(1/2)\sin(m+1/2)$.

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Previously I discussed a variant on the harmonic series, $\sum_{n=1}^\infty\cos n/n$. Last time we showed that

\[ \sum_{n=1}^\infty\frac{\cos n}{n} = \sum_{n=1}^\infty\sum_{m=1}^n\cos m\Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr) , \]

and then showed that the series on the right converges absolutely, by comparison with the series $\sum_{n=1}^\infty3/n^2$.

Since the series on the right converges and the two series have the same value, the series on the left also converges. However, this does not imply that the series on the left also converges absolutely. As a trivial counterexample, if a conditionally convergent series sums to $c$ then $c\sum_{n=1}^\infty \href{http://en.wikipedia.org/wiki/Kronecker_delta}{\delta_{n,1}}$ is an absolutely convergent series which sums to the same value. 🙂

In this post, we answer the question of whether $\sum_{n=1}^\infty\cos n/n$ converges absolutely or not.

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