Consider the series $\sum_{1} ^{\infty} a_n$ where $a_n= \frac{1}{n}$ if $n$ is square and $\frac{1}{n^2}$ otherwise.
How to figure out whether the given series is convergent or not ?
This series is a combination of two series one of which is convergent and the other one is divergent, so I think the whole series should diverge, but i am not sure.
What if $a_n=\frac{1}{n^2}$ if n is odd and $a_n=-\frac{1}{n}$ if n is even?
On
The series with the square criterion converges because the terms $\dfrac1n$ follow the sequence of perfect squares and are sparse. In fact, you are summing $\dfrac1{k^2}$.
The series with the parity criterion diverges because the terms $-\dfrac1n$ are dense (you have an Harmonic series), and they are not compensated by the odd terms, which are known to form a convergent series.
Consider the partial sum $$S_N=\sum_{n=1}^N a_n\ .$$ Any $N$ can be "enclosed" between two squares, say $$M^2\le N<(M+1)^2$$ and we then have $$\eqalign{S_N &=a_1+a_4+\cdots+a_{M^2}+\langle\hbox{all other terms up to $a_N$}\rangle\cr &=1+\frac14+\cdots+\frac1{M^2}+\langle\hbox{all other terms up to $a_N$}\rangle\cr &\le1+\frac14+\cdots+\frac1{M^2}+1+\frac1{2^2}+\cdots+\frac1{N^2}\cr &<2C\ ,\cr}$$ where $$C=\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}6\ .$$ Thus the terms $S_N$ are bounded above; they are clearly increasing since each $a_n$ is positive; and therefore the series is convergent.
In fact, since we have a convergent series of positive terms we can rearrange them at will and evaluate the sum: $$\eqalign{a_1+a_2+a_3+\cdots &=(a_1+a_4+\cdots)+(a_2+a_3+a_5+\cdots)\cr &=\Bigl(1+\frac14+\cdots\Bigr)+\Bigl(\frac14+\frac19+\frac1{25}+\cdots\Bigr)\cr &=\Bigl(1+\frac14+\cdots\Bigr)+\Bigl(1+\frac14+\frac19+\frac1{16}+\cdots\Bigr)\cr &\qquad{}-\Bigr(1+\frac1{16}+\frac1{81}+\cdots\Bigr)\cr &=\frac{\pi^2}6+\frac{\pi^2}6-\frac{\pi^4}{90}\cr &=\frac{\pi^2}3-\frac{\pi^4}{90}\ .\cr}$$