Given:$$\sum_{n=1}^\infty \left( \frac{n^2}{2^n}+\frac{1}{n^2}\right)$$
For testing it's convergence, I have tried using limit form of comparison test in which $\frac{a_n}{a_{n+1}}$ is coming out to be rather complex.
$$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim_{n\to\infty} \left( \frac{2\cdot\left(1+\frac{2^n}{n^4}\right)}{\left(1+\frac{1}{n}\right)^4+\left(\frac{2^{n+1}}{n^4}\right)}+2\cdot\left(\frac{n+1}{n}\right)^2 \right).$$
But I don't know how to simplify further. Kindly guide?
$\dfrac{n^2}{2^n}+\dfrac{1}{n^2} < \dfrac{1}{n^2}+\dfrac{1}{n^2} = \dfrac{2}{n^2}, n > 20$, so it works by comparing it with a known converging series. Note that it can be proven by various ways such as the above ratio test.
The second series is convergent $p$-series $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$ and the first one converges by ratio test $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{2^{n+1}}/\frac{n^2}{2^{n}}=\frac{(n+1)^2}{2n^2}\to\frac{1}{2}<1$$ as $n \to\infty.$ Furthermore, the sum of two convergent series is convergent.
Hint: For $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ the root test shows $$\lim_\infty \sqrt[n]{\frac{n^2}{2^n}}=\frac12<1$$ and the integral test shows $$\sum_{n=1}^\infty \frac{1}{n^2}<\infty$$
You can use the integral test. So that $\sum_{n=1}^\infty \left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)$ converges if and only if $\int_{1}^\infty \left(\frac{x^2}{2^x}+\frac{1}{x^2}\right)dx$. Note that $2^x=e^{-\log(2)x}$. Then, first we compute: \begin{align} \int\frac{x^2}{2^x}dx & = -\frac{x^2}{\log(2)2^x}+\frac{2}{\log(2)}\int \frac{x}{2^x}dx \\ & = -\frac{x^2}{\log(2)2^x}+\frac{2}{\log(2)}\left(-\frac{x}{2^x \log(2)}+\frac{1}{\log(2)}\int \frac{1}{2^x}dx\right) \\ & = -\frac{x^2}{\log(2)2^x}-\frac{2x}{\log(2)^22^x}-\frac{2}{\log(2)^32^x}+C. \end{align} As $\int \frac{1}{x^2}dx=-\frac{1}{x}+C$. We have that: \begin{align} \int_{1}^\infty \left(\frac{x^2}{2^x}+\frac{1}{x^2}\right)dx & = \int_{1}^\infty \frac{x^2}{2^x} dx +\int_{1}^\infty \frac{1}{x^2}dx \\ & = \left(-\frac{x^2}{\log(2)2^x}-\frac{2x}{\log(2)^22^x}-\frac{2}{\log(2)^32^x}-\frac{1}{x}\right)\bigg\rvert_{1}^{\infty} \\ & = \frac{1}{\log(2)^22}+\frac{2}{\log(2)^2}+\frac{2}{\log(2)^32}+1<\infty, \end{align} since $\lim\limits_{x\to\infty}\frac{x^m}{e^x}=0$, for every $m\in\mathbb{N}$. Then, the series converge.
Preliminaries:
$$
\begin{align}
\sum_{k=1}^n\frac1{2^k}
&=\frac12\sum_{k=1}^n\frac1{2^{k-1}}\tag{1a}\\
&=\frac12\sum_{k=0}^{n-1}\frac1{2^k}\tag{1b}\\
&=\frac12+\frac12\sum_{k=1}^{n-1}\frac1{2^k}\tag{1c}\\[6pt]
&=1-\frac1{2^n}\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: pull a factor out front
$\text{(1b)}$: substitute $k\mapsto k+1$
$\text{(1c)}$: pull the $k=0$ term out front
$\text{(1d)}$: subtract the left side of $\text{(1a)}$ from twice $\text{(1c)}$
$$
\begin{align}
\sum_{k=1}^n\frac{k}{2^k}
&=\sum_{k=1}^n\frac1{2^k}+\sum_{k=2}^n\frac{k-1}{2^k}\tag{2a}\\
&=\sum_{k=1}^n\frac1{2^k}+\sum_{k=1}^{n-1}\frac{k}{2^{k+1}}\tag{2b}\\[3pt]
&=2\sum_{k=1}^n\frac1{2^k}-\frac{n}{2^n}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: $k=1+(k-1)$ and $k-1=0$ when $k=1$
$\text{(2b)}$: substitute $k\mapsto k+1$ in the rightmost sum
$\text{(2c)}$: subtract the left side of $\text{(2a)}$ from twice $\text{(2b)}$
$$
\begin{align}
\sum_{k=1}^n\frac1{k^2}
&\le1+\sum_{k=2}^n\frac1{k(k-1)}\tag{3a}\\
&=1+\sum_{k=2}^n\left(\frac1{k-1}-\frac1k\right)\tag{3b}\\
&=1+\left(1-\frac1n\right)\tag{3c}\\[3pt]
&=2-\frac1n\tag{3d}
\end{align}
$$
Explanation:
$\text{(3a)}$: $k^2\ge k(k-1)$
$\text{(3b)}$: Partial Fractions
$\text{(3c)}$: Telescoping Sum
$\text{(3d)}$: algebra
Convergence:
$(1)$ and $(2)$ show that $$ \sum_{k=1}^\infty\frac{k}{2^k}=2\tag{4} $$ $(3)$ shows that $$ \sum_{k=1}^\infty\frac1{k^2}\le2\tag{5} $$ $(4)$ and $(5)$ show that $$ \sum_{k=1}^\infty\left(\frac{k}{2^k}+\frac1{k^2}\right)\le4\tag{6} $$ That is, the series converges.
