Upper bound for truncated taylor series A paper claimed the following but I can't figure out why it's true:
For all $1/2> \delta > 0$, $k\le n^{1/2-\delta}$, and $j\le k-1$ where $n$, $j$, and $k$ are positive integers, the following holds: $$\sum_{i=j}^k \frac{(2/n^{2 \delta})^i}{i!} \le \frac{4}{n^{2\delta j}}.$$
Why is this the case?
Upper-bounding the sum by $(k-j+1)$ times the largest term doesn't seem to work, and it is not obvious to me that replacing $k$ with infinity would work either.  (This paper has many typos in it, so it's also possible that the inequality might not even be true.  The inequality in question is in the proof of Lemma 5.2 of this paper.
 A: Notes:


*

*If I read the notation of the paper correctly (and the putative inequality is on line 16 of page 19), the authors' claim is weaker than in your question, namely
$$
\sum_{i=j}^{k} \frac{(2/n^{2\delta})^{i}}{i!} \leq \frac{4}{n^{\delta j}}
$$
(n.b. $n^{\delta j}$ in the right-hand denominator rather than $n^{2\delta j}$). This doesn't seem to matter, however, since $n^{2\delta}$ can be arbitrarily close to $1$ (if $\delta \ll 1$).

*Since $n$ (and therefore $k$) can be arbitrarily large while $\delta$ can be arbitrarily close to $0$, you can't avoid estimating the infinite tail of the exponential series.

*If $j = 2$, the integers $n$ and $k$ are large, and we let $\delta \to 0$, so in the limit $2/n^{2\delta} \approx 2$, the inequality as stated becomes, approximately,
$$
e^{2} - 1 - 2 = \sum_{i=2}^{\infty} \frac{2^{i}}{i!} \leq 4;
$$
this is false, since $e^{2} - 3 > 4.38$. However, the inequality is true as stated for $j \geq 3$. (The inequality is a straightforward consequence of Taylor's theorem if $j \geq 5$; with a bit more work, one gets $j \geq 3$.)

($j \geq 5$) If $x > 0$ is real and $j$ is a positive integer, Taylor's theorem with the Lagrange form of the remainder applied to $e^{x}$ implies there exists a $z$ with $0 < z < x$ such that
$$
\sum_{i=j}^{\infty} \frac{x^{i}}{i!}
  = e^{x} - \sum_{i=0}^{j-1} \frac{x^{i}}{i!}
  = R_{j}(x)
  = e^{z} \cdot \frac{x^{j}}{j!}.
$$
Setting $x = 2/n^{2\delta}$ and letting $k > j$ be an arbitrary integer, we have
$$
\sum_{i=j}^{k} \frac{(2/n^{2\delta})^{i}}{i!}
  \leq \sum_{i=j}^{\infty} \frac{(2/n^{2\delta})^{i}}{i!}
  = e^{z} \cdot \frac{2^{j}}{j!\, n^{2\delta j}}
  = e^{z} \cdot \frac{2^{j-2}}{j!}\, \frac{4}{n^{2\delta j}}
\tag{1}
$$
for some positive real number $z < 2/n^{2\delta} \leq 2$. Since $e^{2} \approx 7.39$ and $2^{j-2}/j! \leq 1/2$ (with equality if and only if $j = 1$ or $j = 2$), you get the stated inequality within a factor of $e^{2}/2 < 4$ for all $j \geq 1$, and the stated inequality for $j \geq 5$.

($j \geq 3$) The goal is to sharpen the estimate of $e^{z}$. To this end, use the well-known geometric series estimate of the infinite tail:
$$
\sum_{i = j}^{\infty} \frac{x^{i}}{i!}
  = \frac{x^{j}}{j!} \sum_{i = 0}^{\infty} \frac{j!\, x^{i}}{(i+j)!}
  \leq \frac{x^{j}}{j!} \sum_{i = 0}^{\infty} \left(\frac{x}{j + 1}\right)^{i}
  = \frac{x^{j}}{j!} \cdot \frac{1}{1 - x/(j + 1)}.
\tag{2}
$$
(The inequality follows because $j!\, (j + 1)^{i} \leq (i + j)!$ for all positive integers $i$ and $j$.)
Substituting the remainder into (2) gives
$$
e^{z} \cdot \frac{x^{j}}{j!}
  = \sum_{i = j}^{\infty} \frac{x^{i}}{i!}
  \leq \frac{x^{j}}{j!} \cdot \frac{1}{1 - x/(j + 1)},
$$
or
$$
e^{z} \leq \frac{1}{1 - x/(j + 1)}
  = \frac{j + 1}{j + 1 - x}.
$$
If $j \geq 3$, then (since $x < 2$) we have $e^{z} \leq (j + 1)/(j - 1) \leq 2$. Inequality (1) therefore implies
$$
\sum_{i=j}^{k} \frac{(2/n^{2\delta})^{i}}{i!} \leq \frac{4}{n^{2\delta j}}.
$$
