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I am having trouble in proving the pointwise convergence of $$ g(x)=\sum_{n=1}^\infty \frac{\sin(\sqrt{n}x)}{n}$$ for all real numbers $x$ using elementary methods (e.g. integral test, Weierstrass M-test, Abel's Test, Dirichlet's Test, Comparison with Riemann sum, etc.).

Can someone help me on this?

Thanks.

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6 Answers 6

up vote 5 down vote accepted

This may be very closely related to Robert's suggestion.

Elaborated answer

Consider the following inequality (with $n,m \to \infty$): $$ \Bigg|\sum\limits_{k = n}^m {\frac{{\sin (\sqrt k x)}}{k}} - \sum\limits_{k = n}^m {\int_k^{k + 1} {\frac{{\sin (\sqrt u x)}}{u}du} } \Bigg| \le \sum\limits_{k = n}^m {\int_k^{k + 1} \Bigg| \frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg|du} . $$ For fixed $x > 0$, show that, for any $u \in [k,k+1]$, $$ {\Bigg|\frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg|} \leq \frac{x}{{2k\sqrt k }} + \frac{1}{{k^2 }}. $$ (Thus the same inequality holds for the integral from $k$ to $k+1$ of the left-hand side.) For this purpose, first write $$ \Bigg|\frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg| = \Bigg|\frac{{\sin (\sqrt k x)k - \sin (\sqrt u x)k + \sin (\sqrt k x)(u - k)}}{{ku}}\Bigg|. $$ Then apply the triangle inequality, and use the mean value theorem (twice). Further note that $\int_1^\infty {\frac{{\sin (\sqrt u x)}}{u}} du$ converges (for this purpose, make a change of variable $y=\sqrt u x$). The rest is straightforward.

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Look at the sequence of partial sums up to each sign change in the sine, i.e. combine all consecutive terms with the same sign. Since the length of these groups of terms grows as $\sqrt{n}$, their sum decays as $1/\sqrt{n}$, so you get an alternating series with terms converging to zero.

Edit: Didier rightly pointed out that I didn't show that the absolute values of the terms strictly decrease. To show this rigorously would probably be more hassle than pursuing one of the approaches in the other answers, but here's a not-so-rigorous argument: Each group of terms can be considered as an approximation with the trapezoidal rule of the integral over $\sin \sqrt{ux}/u$ over an interval of $\pi/x$, scaled in proportion to the number of terms in the group. The scaling by the number of terms grows as $\sqrt{n}$ and the integrand falls off as $1/n$, for a decay of $1/\sqrt{n}$. There are three sources of approximation error. Two occur at the endpoints, one from the fact that the trapezoidal rule specifies a weight of $1/2$ there, and another from the fact that the ends of the sum and the integral don't coincide. Both of these errors are proportional to the outermost terms in the group, which are of order $1/\sqrt{n}^3$, and to the width of the integration interval they represent, which is of order $1/\sqrt{n}$, for a total decay of $1/n^2$. The third approximation error is the error in the trapezoidal rule itself, which is cubic in the width of the integration intervals for a factor of $1/\sqrt{n}^3$, and proportional to the second derivative of the integrand, which falls off with $1/n$, for a total decay of $1/\sqrt{n}^5$. So both approximation errors decay faster than the difference of order $1/\sqrt{n}^3$ in the contributions of order $1/\sqrt{n}$.

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The alternating property you invoke might not be so easy to prove. That is, the fact that the sequence of absolute values is nonincreasing. And without this property, I do not know how to conclude. –  Did Mar 15 '11 at 9:53
    
@Didier Piau: There may be a misunderstanding here. By "combining all consecutive terms with the same sign", I meant to group them together and consider the series whose terms are the sums within these groups. This is alternating by construction, since each term is a sum of terms with the same sign. If that isn't clear from what I've written, please suggest how I might improve the wording. –  joriki Mar 15 '11 at 10:02
    
The trouble is that alternating series usually means that the terms of the series have alternating signs AND nonincreasing absolute values, and that the result you alluded to becomes false if one forgets the latter condition. So, leaving terminology questions aside, your method definitely needs the absolute values of the finite sums you consider to be nonincreasing--and it is not obvious from what you wrote that they are. –  Did Mar 15 '11 at 10:15
    
@Didier Piau: You're absolutely right -- I edited the answer in response, but it's no longer any simpler now than any of the other approaches, probably to the contrary... Thanks for spotting the omission. –  joriki Mar 15 '11 at 15:47

I would try approximating $\sin(\sqrt{n} x)/n$ by $\int_{\sqrt{n-1/2}}^{\sqrt{n+1/2}} \sin(t x)/(2 t) \, dt$, noting that $\int_{0}^\infty \sin(t)/t \, dt$ converges.

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You can use summation by parts here: If $(a_n)_{n \geq 1}$ is a sequence of complex numbers, and $f : [1, \infty] \longrightarrow {\bf C}$ is a $C^1$ function then $$ \sum_{n \leq N} a_n f(n) = (\sum_{n \leq N} a_n) f(N) - \int_{1}^{N} (\sum_{n \leq t} a_n) f'(t) \; d \, t.$$ Applying this to $a_n = {1 \over n}$ and $f(t) = \sin(\sqrt{t}x)$, one gets $$\sum_{n \leq N} {\sin(\sqrt{n}x) \over n} = H(N)\sin(\sqrt{N}x) - \int_{1}^{N} H(\lfloor t \rfloor) {x\cos(\sqrt{t}x) \over 2\sqrt{t}} \; d \, t$$ Here $H(n)$ is the $n$th harmonic number $1 + {1 \over 2} + ... + {1 \over n}$. Fortunately, there are good asymptotics for $H(n)$, namely that $H(n) = \ln(n) + \gamma + O(1/n)$. As a result, the $H(\lfloor t \rfloor)$ appearing in the above integral differs from $\ln(t) + \gamma$ by at most ${C \over t}$. Since ${\displaystyle \int_1^{\infty} {1 \over t} {x\cos(\sqrt{t}x) \over 2\sqrt{t}}\,dt}$ is absolutely integrable, one can replace $H(\lfloor t \rfloor)$ by $\ln(t) + \gamma$ in that integral when trying to prove convergence. One can similarly disregard the $O({1 \over N})$ term in the $H(N)$ on the left. So our goal is to show that the following converges as $N$ goes to infinity. $$(\ln(N) + \gamma) \sin(\sqrt{N}x) - \int_{1}^{N} (\ln(t) + \gamma){x\cos(\sqrt{t}x) \over 2\sqrt{t}} \; d \, t.$$ We now integrate by parts again, in the opposite direction of our original summation by parts; we differentiate $(\ln(t) + \gamma)$ and integrate ${x\cos(\sqrt{t}x) \over 2\sqrt{t}}$ back to $\sin(\sqrt{t} x)$. Our goal now becomes to show the finiteness of $$\int_1^{\infty}{\sin(\sqrt{t} x) \over t}\,dt$$ As others have said above, by changing $t$ to $t^2$, this is equivalent to showing the convergence of ${\displaystyle \int_1^{\infty} {\sin(t x) \over t}\,dt}$, which in turn by replacing $t$ by $tx$ is equivlalent to showing the convergence of ${\displaystyle \int_x^{\infty} {\sin(t) \over t}\,dt}$ which is well-known.

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Nice -- I'd accept this one if it were my question :-) –  joriki Mar 15 '11 at 21:56

Here is the elaboration. Each sum of the same sign, $$ S(m)= \sum\limits_{m<\sqrt{n}x\leq m+1}\frac{1}{n} \sin(\sqrt{n}x \pi), $$ has its absolute value sandwiched between $$ \left(\frac{x}{m}\right)^2 |D(m)|\quad \text{and}\quad\left(\frac{x}{m+1}\right)^2|D(m)|, $$ where $$ D(m)=\sum\limits_{m<\sqrt{n}x\leq m+1}\sin(\sqrt{n}x \pi). $$ Replace summation by $0$ if summation is empty. Take a little care with $m = 0$. We have $x^2 D(0) \leq S(0)\leq D(0)$, and $S(0)$ is non empty if $0< x\leq 1$. Then estimate $|D(m)|$ by using $h(t)= \sin (\sqrt{t} x \pi)$ to lie between $2(2m+1)/(\pi x^2) +1$ and $2(2m+1)/(\pi x^2) -1$. Now $D(m)\geq 0$ when $m$ is even and not greater than $0$ when $m$ is odd. Then you will find that series $$ \sum\limits_{m=0}^{\infty} S(m)=\sum\limits_{n=1}^\infty\frac{\sin(\sqrt{n}x\pi)}{n} $$ is sandwiched between two convergent series. Note that the number of terms in each summation $S(m)$ depends on $x$ and tends to infinity as $x$ tends to $0$ on the right. Note that $|S(m)|$ tends to $0$ as $m$ tends to infinity. Thus the convergence of $S(m)$ implies the convergence of original series. This of course implies the convergence of for $x > 0$ and hence for $x < 0$ and convergence at $0$ is obvious.

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Group terms with same sign together and consider the series whose terms are the sums within these groups. Make an estimate of each term, by using the property that sine is increasing over half of $\pi$ and decreasing after that. Not sine directly but via the function $\sin(\sqrt{t}x)$, $x$ is fixed. Each term is sum over the integer $n$ such that $\frac{\sin(\sqrt{n}x)}{n}$ is of the sme sign. Then you will find that this series is sandwiched between two convergent series, which are sum of convergent $p$ series and alternating series of constant terms. And so it is convergent. This can be applied to the series $\sum\frac{\sin(\sqrt{n}x)}{n^a}$, with $1/2< a < 1$.

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Perhaps you would care to elaborate? Also, by putting formula between dollar signs, you can use a subset of latex capability. For instance $x^2 + y^2 = r^2$ looks like $x^2 + y^2 = r^2$. –  Aryabhata Jun 23 '11 at 19:10

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