Examine the convergence of the series: sum from n=1

Examine the convergence of the series: $$ \sum_{n=1}^{\infty} \sqrt[n]{|x|^{n^2}+|y|^{n^2}}. $$

Step 1 · Apply the root test

With $a_n=\sqrt[n]{|x|^{n^2}+|y|^{n^2}}$:

$$ \sqrt[n]{a_n} = \left(|x|^{n^2}+|y|^{n^2}\right)^{1/n^2}. $$

Step 2 · Factor and evaluate the limit

Assume WLOG $|x|\ge|y|$. Factor out $|x|^{n^2}$:

$$ \left(|x|^{n^2}+|y|^{n^2}\right)^{1/n^2} = |x|\cdot\left(1+\left(\frac{|y|}{|x|}\right)^{n^2}\right)^{1/n^2}. $$

If $|x|>|y|$, then $\left(\frac{|y|}{|x|}\right)^{n^2}\to0$ rapidly, and since $(1+t_n)^{1/n^2}=e^{\ln(1+t_n)/n^2}\to e^0=1$ for $t_n\to0$, we get $\rho=|x|$. If $|x|=|y|$, then $(2|x|^{n^2})^{1/n^2}=2^{1/n^2}|x|\to|x|$. In general:

$$ \rho = \limsup_{n\to\infty}\sqrt[n]{a_n} = \max\{|x|,|y|\}. $$

Step 3 · Apply the convergence criterion

Converges if $\max\{|x|,|y|\}<1$; diverges if $\max\{|x|,|y|\}>1$. At $\max\{|x|,|y|\}=1$ (say $|x|=1$), $a_n=\sqrt[n]{1+|y|^{n^2}}\ge\sqrt[n]1=1$, so $a_n\not\to0$: the series diverges here too.

Summary Table

Additional Notes

• The structure closely mirrors Problem 77's $\sqrt[n]{a^n+b^n}\to\max\{a,b\}$, but nested one level deeper (an outer $n$-th root of an already-rooted quantity), which is why the exponent ends up as $1/n^2$ rather than $1/n$.
• Both variables $x$ and $y$ play a genuinely symmetric role: convergence depends only on whichever of $|x|,|y|$ is larger, so this is really a two-variable generalization of the single-variable geometric series threshold.
• The boundary case $\max\{|x|,|y|\}=1$ needed a direct term-by-term check (rather than the root test alone, which is inconclusive at $\rho=1$) — the same pattern seen in Problems 75 and 76.