Converging Series: Comparison Test w/ 1/n^2

[REVIANNA]

Mod note: Moved from Homework section
I know that $1/n^4$ converges because of comparison test with $1/n^2$ (larger series) converges .
how do I know $1/n^2$ converges?
coz I cannot compare it with $1/n$ harmonic series as it diverges.

@REVIANNA, if you post in the Homework & Coursework sections, you must include an attempt. Your question seemed like more of a general question, but not a homework problem, so I moved your thread.

 

[Samy_A]

Hint: for $n\geq 2$: $\frac{1}{n²}\leq \frac{1}{n²-n}=\frac{1}{n(n-1)}$

 

[Ray Vickson]

Homework Statement

I know that $1/n^4$ converges because of comparison test with $1/n^2$ (larger series) converges .
how do I know $1/n^2$ converges?
coz I cannot compare it with $1/n$ harmonic series as it diverges.

What is "coz"?

Anyway, the slickest way to show that $\sum 1/n^p$ converges if $p > 1$ and diverges if $p = 1$ (or $p < 1$) is to note that for $n \geq 2$ we have
$$\int_n^{n+1} \frac{dx}{x^p} < \frac{1}{n^p} < \int_{n-1}^n \frac{dx}{x^p}$$
so you can get easily-computed upper and lower bounds on $\sum_{n=2}^N 1/n^p$.