Find lim as p to infinity sum from k=1 to infinity
Find $\displaystyle\lim_{p\to\infty}\sum_{k=1}^{\infty}\frac1{k^{2p}}$.
Explanation of Method
The $k=1$ term is always exactly $1$; every other term $1/k^{2p}$ ($k\ge2$) shrinks to $0$ as $p\to\infty$ since $k^2\ge4$. Bounding the tail $\sum_{k\ge2}1/k^{2p}$ by a convergent integral shows this tail vanishes as $p\to\infty$, isolating the answer as exactly the $k=1$ term.
- Isolate the $k=1$ term, which is identically $1$.
- Bound the tail $\sum_{k\ge2}1/k^{2p}$ via the integral test.
- Take the limit of the tail bound to show it vanishes.
Step 1 · Split off the first term
$$ \sum_{k=1}^{\infty}\frac1{k^{2p}} = 1 + \sum_{k=2}^{\infty}\frac1{k^{2p}}. $$
Step 2 · Bound the tail via the integral test
For $p\ge1$, $f(u)=1/u^{2p}$ is positive, continuous, and decreasing for $u\ge1$, so $$ \sum_{k=2}^{\infty}\frac1{k^{2p}} < \int_1^{\infty}\frac{du}{u^{2p}} = \left[\frac{u^{1-2p}}{1-2p}\right]_1^{\infty} = \frac1{2p-1}. $$
Step 3 · Take the limit
Since $0\le\sum_{k=2}^{\infty}1/k^{2p}<\dfrac1{2p-1}\to0$ as $p\to\infty$, the squeeze theorem gives $\displaystyle\lim_{p\to\infty}\sum_{k=2}^{\infty}\frac1{k^{2p}}=0$, and therefore $$ \lim_{p\to\infty}\sum_{k=1}^{\infty}\frac1{k^{2p}} = 1 + 0 = 1. $$
Summary Table
| Quantity | Value |
|---|---|
| $k=1$ term | $1$ (constant in $p$) |
| Tail bound | $\sum_{k\ge2}1/k^{2p}<1/(2p-1)$ |
| $\lim_{p\to\infty}\zeta(2p)$ | $1$ |
Additional Notes
- This confirms the intuitive picture: as $p\to\infty$, the Riemann zeta function $\zeta(2p)$ approaches its "trivial" limit of $1$, since every term beyond $k=1$ is crushed exponentially fast by the growing exponent.
- Compare with Problem 199, where $\zeta(x)\to\infty$ as $x\to1^+$ (the opposite extreme) — together these two limits sketch the full qualitative shape of $\zeta$ on $(1,\infty)$: unbounded near $1$, decreasing, and asymptotically flattening to $1$ as $x\to\infty$.
- The same isolate-the-dominant-term technique appears in Problem 214, where the $n=0$ term of that series similarly survives a limit while the geometric tail collapses to a computable constant.