True/ False: If $\sum_{k=1}^{\infty}f(k)$ converges, then $f(x)\rightarrow0$ as $x\rightarrow\infty$

Question

The solutions in my textbook say this is false, but I do not see how this is any difference from if $\sum_{k=1}^{\infty}a_{k}$ converges, then $a_{k}\rightarrow0$ which is true in the solutions. $f(x)$ may have some discontinuities on some non-natural values of $x$, but shouldn't that not effect the limit's long-term behaviour towards infinity? Am I missing something?

Answer 1

Consider the function $f$ with $f(x) = 0$ for all rational $x$ and $f(x) = 1$ otherwise.

If you want a continuous example, consider $f$ such that $f(x) = 0$ for all integers $x$, $f(x) = 1$ for all half-integers $x$ and $f(x)$ is linearly interpolated between the nearest integer and half-integer for all other $x$.

Answer 2

The problem as phrased is false; consider the function $$ f(x) = \begin{cases} 0 & x \ \text{is an integer} \\ x & \text{otherwise}. \end{cases} $$ Then $\sum_{k = 1}^\infty f(k) = 0$ since $f(k)$ is zero for each integer $k$, but $f(x)$ does not have a limit towards infinity.

Answer 3

For a smooth function, how about $$f(x) = x\cdot \sin(\pi x)\quad ?$$ For any integer $k$, $f(k) = 0$, but $f$ grows without bound (oscillating positive and negative) as $x$ increases.

And if you think that the positives and negatives "average out" to a limit of 0 (which is incorrect anyway), just take $$g(x) = x\cdot\sin^2(\pi x)$$ which is always non-negative if $x$ is.