Does this sum converge or diverge?

Question

Does the infinite sum $\large{\sum_{n=1}^\infty \frac{1}{n^{x_{\small{n}}}}}$ converge if $x_n$ is a random variable (generated within each term) that takes values between $0$ and $2$ with equal probability converge or diverge? I have a suspicion that it diverges, but I don't know how to prove it.

Answer 1

Since the series in question is based on random values, it does not make sense to ask if it converges; rather one should ask about the probability that it will converge to a finite value.

Define the Boolean function $$ b(x) = \begin{cases} 0, \text{if $x \geq 1$} \\ 1, \text{if $x \lt 1$} \end{cases} $$

Now set $b_n = b(x_n)$. Clearly, $0 \lt 1/n^{x_n}$, and if $x_n < 1$, then $1/n < 1/n^{x_n}$. Thus it must always be the case that $$b_n / n \lt 1/n^{x_n}$$

Likewise, the sum: $$ \sum_{n=1}^\infty {b_n\over n} \lt \sum_{n=1}^\infty {1\over n^{x_n}} $$

Since each $x_n$ is randomly (evenly) distributed over $0 \lt x_n \lt 2$, the $b_n$ are distrbuted evenly as $0$ or $1$.

(Edited here to be more formal)

Now, define a new series $c_m$ with $c_0 = b_1$ and $$ c_m = \sum_{n=2^{m-1}+1}^{2^{m}} \frac{b_n}{n} (m \geq 1) $$ This is just separating the $b_n/n$ terms into groups. Each $c_m$ is a sum of $2^{m-1}$ terms (except for $c_0$, which has $1$ term, not $\frac12$). Since every $b_n/n$ is part of exactly one $c_m$, we must have: $$ \sum_{m=0}^\infty c_m = \sum_{n=1}^\infty {b_n\over n} $$

Each $c_m$ is a sum of $2^{m-1}$ terms, and each term has a denominator less than $2^{m}$. If all the $b_n$ are $1$, then $c_m > 1/2$. On average, half the $b_n$ are $1$, so the average value for each $c_m > 1/4$. Specifically, $c_m < 1/4$ only if more than half of the related $b_n$ are $0$. Since each $b_n$ is independently $0$ or $1$ with $50\%$ probability, the probability that more than half of them are $0$ is clearly less than $1/2$.

The sum $$ \sum_{m=1}^\infty c_m $$ is a sum of an infinite number of terms, each of which is $>1/4$ with $>50\%$ probability. This series clearly diverges. More formally, it should be straightforward to demonstrate: $$ \forall (T>0) \forall (\epsilon>0): \exists (N) \mid P\left(\sum_{m=0}^N c_m < T\right) < \epsilon $$ That is one definition of "divergent" for a probabilistic series.

From above: $$ \sum_{n=1}^\infty {1\over n^{x_n}} > \sum_{n=1}^\infty {b_n\over n} = \sum_{m=0}^\infty c_m $$

The last term diverges, so the others must also.

Answer 2

NOTE: At the time this post was written, the problem was $\sum_n \frac{1}{n^x}$ - the asker has revised their problem statement.

With probability $\frac{1}{2}$, the series diverges (it is a $p$-series with $p\leq 1$ for $x \in [0,1]$) and with probability $\frac{1}{2}$ the series converges (it is a $p$ series with $p>1$ for $x \in (1,2]$).

Note the random variable $x$ is fixed over the sum, so it just becomes a calc 1 convergence problem for each value $x$ takes.

Answer 3

I don't know if I exactly have a proof, but here's a thought. The infinite series of reciprocals of the prime numbers diverges. How likely is it that $ \ x_n \ > \ 1 \ $ "often enough" to produce a series with terms that can bring the series to convergence despite that? That is, can there be a high enough "density" of terms that make the series convergent against the sum of terms that would cause divergence? Perhaps there is an argument something like comparing $ \ \sum_{n=1}^{\infty} \ \frac{1}{n^{x_n}} \ $ to $ \ \sum_{n=1}^{\infty} \ \frac{1}{p_n} \ $ , which has a lower "density" of terms than, say, the harmonic series. (I suspect the probability of having a convergent series is essentially zero.)

Answer 4

If $P(x_n=0)=c$ for every $n$ with $c\gt0$ and the sequence $(x_n)$ is independent, then the random set $Z=\{n\geqslant1\mid x_n=0\}$ is almost surely infinite and $$ \sum_{n\geqslant1}\frac1{n^{x_n}}\geqslant|Z|, $$ hence the series on the LHS diverges almost surely.

The same result holds, assuming only independence and that $P(x_n\leqslant1)\geqslant c$ for every $n$, for some $c\gt0$, but the proof is slightly more elaborate.

To see how the proof works, assume as in the question that $(x_n)$ is i.i.d. and uniformly distributed on the interval $(0,2)$ and consider, for every $n$, the random set $$L_n=\{k\mid 2^n\leqslant k\lt2^{n+1}, x_k\leqslant1\},$$ of size $\ell_n=|L_n|$, and the part $S_n$ of the series restricted to $L_n$, that is, $$ S_n=\sum_{k=2^n}^{2^{n+1}-1}\frac1{k^{x_k}}. $$ For every $k$ in $L_n$, $k^{x_k}\leqslant k\leqslant2^{n+1}$ hence $$ S_n\geqslant\sum_{k\in L_n}\frac1{k}\geqslant\frac{\ell_n}{2^{n+1}}. $$ For every $k$, let $U_k=\mathbf 1_{x_k\leqslant1}$, then $(U_k)$ is i.i.d. Bernoulli with parameter $\frac12$ and $$\ell_n=\sum\limits_{k=2^n}^{2^{n+1}-1}U_k $$ is binomial $(2^n,\frac12)$ with mean $E(\ell_n)=2^n\cdot\frac12$ and variance $\sigma^2(\ell_n)=2^n\cdot\frac14$ hence $$ [\ell_n\leqslant2^{n-2}]\subset[|\ell_n-E(\ell_n)|\geqslant2^{n-2}], $$ and Bienaymé-Chebychev inequality yields $$ P(\ell_n\leqslant2^{n-2})\leqslant\frac{\sigma^2(\ell_n)}{(2^{n-2})^2}=\frac4{2^n}. $$ The RHS is summable hence Borel-Cantelli lemma implies that, almost surely, for every $n$ large enough, $\ell_n\gt2^{n-2}$. When $\ell_n\gt2^{n-2}$, $$ \sum_{k=2^n}^{2^{n+1}-1}\frac1{k^{x_k}}\geqslant\frac18. $$ The RHS does not converge to $0$. This implies the almost sure divergence of the full series $$ \sum_{k=1}^{\infty}\frac1{k^{x_k}}. $$ The proof when one assumes only that $(x_n)$ is independent and that $P(x_n\leqslant1)\geqslant c$ for every $n$, for some $c\gt0$, is quite similar.