Vtyjkyuk

Suppose f is continuous on $[0,1]$. Find

$$L=lim_nrightarrow infty frac1n sum_k=1^n (-1)^k f(frackn) $$

My attempts : I was thinking about Riemann sum... but $(-1)^k$ creates confusion.

Thanks in advance.

I am here to give an analogous example. Maybe this method is also applicable to your question.

Compute $sum_1^infty (-1)^n-1 1/n$.

If we have some knowledge about harmonic series $sum_1^infty 1/n$:
$$
1 +frac 12 + frac 13 + cdots + frac 1n = log(n) + gamma + o(1) quad [n to infty],
$$
where $gamma$ is the Euler constant, then we could rewrite
$$
1-frac 12 + frac 13 - frac 14 + cdots + (-1)^n-1 frac 1n = sum_1^n frac 1j - 2 sum_1^lfloor n/2rfloor frac 1k,
$$
and use the asymptotic estimate above to get the result.

UPDATE

Solution. $blacktriangleleft$ Let $s_n = sum_1^n (-1)^k-1 f(k/n)$. Then $-L = lim s_n$ if $s_n$ converges.

Consider $s_2n$. By the similar operation above,
$$
s_2n = frac 12nleft(fleft( frac 12nright) + fleft(frac 22n right)+ cdots + fleft(frac 2n2n right) right) - 2 cdot frac 1 2n left( f left( frac 1nright)+ f left(frac 2nright) + cdots + fleft(frac nnright) right).
$$
Since $f in mathcal C[0,1]$, $f in mathcal R[0,1]$. Then
$$
lim s_2n = lim_n frac 12n sum_1^2n fleft( frac j2nright) - lim_n frac 1n sum_1^n f left(frac jn right) = int_0^1 f - int_0^1 f = 0.
$$
Now for $s_2n+1$,
$$
s_2n+1 = frac 1 2n+1 sum_1^2n+1 f left( frac j2n+1 right) - frac 2 2n+1 sum_1^n f left( frac 2j2n+1right).
$$
Clearly,
$$
frac j-1n leqslant frac 2j2n+1 = frac j n+1/2 leqslant frac jn,
$$
where the first inequality holds because of
$$
frac j-1n leqslant frac 2j2n+1 iff (j-1)(2n+1) leqslant 2jn iff -(2n+1) + j leqslant 0 quad [j leqslant n].
$$
Thus $(1/n)sum_1^n f(2j/(2n+1))$ is a Riemann sum w.r.t. to the partition $0, 1/n, 2/n, ldots, 1$. Then
$$
lim s_2n+1 =lim frac 12n+1 sum_1^2n+1 f left( frac j 2n+1right) - lim frac 2n2n+1 cdot lim frac 1n sum_1^n fleft( frac 2j2n+1right) = int_0^1 f - int_0^1 f = 0.
$$
Therefore $s_n$ is a sequence that $lim s_2n = lim s_2n+1 = 0$. Conclusively, $L = 0$. $blacktriangleright$

Seems that $f in mathcal R [0,1]$ might be sufficient.

Contrary to your name, you are not stupid.
You possess a very important ability in problem solving: realizing when you took the wrong approach. In such a case, you just have to abandon that road, and think about a different one.

Here, continuity is the key. Moreover, continuity on a bounded, closed interval, which implies uniform continuity. So if $n$ is large, the values $f(k/n)$ and $f((k+1)/n)$ are very close to each other.

I hope this hint helps. Mind the borders! What if $n$ is odd? What if $n$ is even?

If $f$ is continuous on $[0,1]$ it is uniformly continuous; for any $ngeq 1$ let $$Delta_n = sup_substackx-yleft|f(x)-f(y)right|$$
and $M=sup_xin[0,1]left|f(x)right|$. By the triangle inequality
$$ left|sum_k=1^n(-1)^k fleft(tfracknright)right| leq M + fracn2Delta_n $$
and since $Delta_n to 0$ as $nto +infty$, the wanted limit is zero.