If $\sum_{i=0}^{\infty} a_i$ converges, does $\sum_{i=0}^{\infty} \arcsin(a_i)$ converge?

Question

If $\sum_{i=0}^{\infty} a_i$ converges, does $\sum_{i=0}^{\infty} \arcsin(a_i)$ converge? If not, give a counterexample.

Answer 1

Note that $\arcsin(2x)-2\arcsin(x)>0$ for $x>0$. Now we can easily produce a converging $\sum a_i$ with the property that $a_{3k-1}=a_{3k-2}=-\frac12a_{3k}$ for all $k$ ; in fact it suffices to ensure $a_{3k}\to 0$ to have convergence of the series. At the same time $\arcsin a_{3k-2}+\arcsin a_{3k-1}+\arcsin a_{3k}$ can be strictly positive and we can repeat the same triple / the same positive contribution, until these repeated triples contribute at least $\frac 1n$, say, to the arcsine sum. Then the arcsine series compares to the harmonic series and diverges.

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More generally, we have

Proposition. Assume $f\colon D\to \Bbb R$ is a function defined in a neighbourhood of $0$ and such that for all $\epsilon>0$ there exist $x,y,z\in(-\epsilon,\epsilon)\cap D$ and $x+y+z=0$ and $f(x)+f(y)+f(z)\ne 0$. Then there exists $(a_i)_{i\in \Bbb N}$ with $a_i\in D$ for all $i$ and $\sum a_i=0$ and $\sum f(a_i)$ diverges.

Proof. For $m\in\Bbb N$ pick $x_m,y_m,z_m\in(-\frac1m,\frac1m)\cap D$ with $x_m+y_m+z_m=0$ and $f(x_m)+f(y_m)+f(z_m)\ne 0$. Wlog. $f(x_m)+f(y_m)+f(z_m)> 0$ for infinitely many $m$. Hence wlog. $f(x_m)+f(y_m)+f(z_m)> 0$ for all $m$. Let $r_m=\lceil\frac1{f(x_m)+f(y_m)+f(z_m)}\rceil$ and $s_m=r_1+\ldots+r_m$. Then for $n\in \Bbb N$, $n=3k-t$, $t\in\{0,1,2\}$ let $$\tag1a_n=\begin{cases}x_m&\text{if }t=0\\y_m&\text{if }t=1\\z_m&\text{if }t=2\end{cases}\quad\text{where $m$ is minimal with $s_m\ge k$}.$$ Then the $n$th partial sum does not exceed $\frac 1m$ in absolute value (with $m$ as defined in $(1)$), hence $\sum a_i$ converges to $0$. On the other hand, $$\sum_{i=3s_m-2}^{3s_{m+1}-3}f(a_i) =r_m\cdot(f(x_m)+f(y_m)+f(z_m))\ge 1$$ so that $\sum_{i=3s_1-2}^{3s_{m+1}-3}f(a_i)\ge m$ and $\sum_{i=1}^{\infty}f(a_i)$ diverges. $_\square$

Remark: For continuous $f$, the condition of the proposition is already given if there is no neighbourhood of $0$ on which $f$ is linear. In particular, the condition holds for $f(x)=\arcsin x$.

Answer 2

If $\;\sum_i a_i$ converges absolutely, $\;\arcsin a_i$ converges absolutely.

Indeed, $a_i$ tends to $0\;$ and $\;a_i\sim \arcsin a_i$. Two series with equivalent terms both converge absolutely or both diverge absolutely.