Find lim as x to infinity sum from n=0 to infinity
Find $\displaystyle\lim_{x\to\infty}\sum_{n=0}^{\infty}\frac{x^2}{1+2^nx^2}$.
Explanation of Method
Each term individually tends to $1/2^n$ as $x\to\infty$. A single $x$-independent bound $1/2^n$ (valid for every $x\ne0$, not just large $x$) dominates the whole series and is itself summable, so the limit can be exchanged with the sum via dominated convergence.
- Find the pointwise limit of each term as $x\to\infty$.
- Bound every term by $1/2^n$, uniformly for all $x\ne0$.
- Exchange limit and sum using this dominating, summable bound.
Step 1 · Term-wise limit
For fixed $n\ge0$: $$ \frac{x^2}{1+2^nx^2} = \frac1{\frac1{x^2}+2^n} \xrightarrow[x\to\infty]{} \frac1{2^n}. $$
Step 2 · A uniform dominating bound
For every $x\ne0$ and $n\ge0$: dropping the $1$ in the denominator, $$ \frac{x^2}{1+2^nx^2} \le \frac{x^2}{2^nx^2} = \frac1{2^n}. $$ This bound holds for all $x\ne0$ (not just large $x$), and $\sum_{n=0}^{\infty}1/2^n=2<\infty$.
Step 3 · Exchange limit and sum
Since the terms are dominated by the summable, $x$-independent sequence $1/2^n$, the limit may be taken term by term (dominated convergence): $$ \lim_{x\to\infty}\sum_{n=0}^{\infty}\frac{x^2}{1+2^nx^2} = \sum_{n=0}^{\infty}\lim_{x\to\infty}\frac{x^2}{1+2^nx^2} = \sum_{n=0}^{\infty}\frac1{2^n} = \frac1{1-\frac12} = 2. $$
Summary Table
| Quantity | Value |
|---|---|
| Term-wise limit | $1/2^n$ |
| Dominating bound | $1/2^n$, $\sum=2$ |
| Final limit | $2$ |
Additional Notes
- The bound $x^2/(1+2^nx^2)\le1/2^n$ is exact and $x$-independent for every nonzero $x$ — no restriction to $x\ge1$ is even needed, which simplifies the exchange-of-limit justification considerably compared to a naive tail-splitting approach.
- This is the same dominated-convergence idea used implicitly in Problems 189–191's asymptotic comparisons: a uniform, summable bound lets termwise limiting behavior determine the limit of the whole series.
- Compare with Problem 215, where an analogous limit-exchange argument (again justified by a uniform tail bound) shows $\zeta(2p)\to1$ as $p\to\infty$ — both problems isolate a single dominant term ($n=0$ here, $k=1$ there) that survives the limit while the rest of the tail vanishes.