A sequence $(\phi_n)$ is defined as follows : $$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$$ Show that the sequence is convergent.
Because this sequence is monotonic, proving it is bounded above will be sufficient to prove that it is convergent.
So how to show that this sequence is bounded above?
Here's a simple approach :
$$k^4 \geq k^2 \geq k(k-1)$$ so :
$$\phi_n=1+\sum_{k=2}^{n} \frac{1}{k^4} \leq 1+\sum_{k=2}^{n} \frac {1}{k(k-1)}$$ but the later sum telescopes because $$\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$$ so :
$$\phi_n \leq 1+1-\frac{1}{n}=2-\frac{1}{n}<2$$ so this sequence is bounded and then because of monotony it's also convergent .
As an aside the sum actually converges to :$$\zeta(4)=\frac{\pi^4}{90}$$
Since $t \mapsto \dfrac{1}{t^4}$ is decreasing, we have $$\frac{1}{(k+1)^4}=\int_{k}^{k+1}\frac{dt}{(k+1)^4}\leq \int_{k}^{k+1}\frac{dt}{t^4},\,\,\, k\geq1,$$ which gives$$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}=1+\sum_{k=1}^{n-1}\frac{1}{(k+1)^4}\leq 1+\int_{1}^{n}\frac{dt}{t^4}=\frac{4}{3}-\frac{1}{3n^3}<\frac{4}{3}.$$
The following approach will also give a reasonable bound:
$$1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+\frac{1}{8^4}+\frac{1}{9^4}+\ldots+\frac{1}{n^4} \\ < 1+\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{8^4}+\frac{1}{8^4}+\ldots+\frac{1}{n^4} \\ < 1+\frac{2}{2^4}+\frac{4}{4^4}+\frac{8}{8^4}+\frac{16}{16^4}+\ldots = 1+\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{8^3}+\frac{1}{16^3}+\ldots \\ = 1+\frac{1}{8}+\frac{1}{8^2}+\frac{1}{8^3}+\frac{1}{8^4}+\ldots < \sum_{k=1}^{\infty} \frac{1}{8^k} = \frac87$$
$$ \phi_n = \sum_{k=1}^n \frac{1}{k^4} \leq \sum_{k=1}^n \frac{1}{k^2} \leq \sum_{k=1}^{\infty} \frac{1}{k^2} < \infty $$
