I must use the definition of limit to prove that the sequence $$\left(\frac{ni}{2^n}\right)_{n\in\mathbb{N}}$$ converges.
What I did was to set out the definition of limit: $$\lim _{n\to\infty}\left(\frac{ni}{2^n}\right)=0\Leftrightarrow[\forall\epsilon>0,\exists n_0>0/\forall n:(n\in\mathbb{N} \wedge n>n_0\Rightarrow|\left(\frac{ni}{2^n}\right)-0|<\epsilon)]$$ Now I must prove: $$\left|\left(\frac{ni}{2^n}\right)\right|<\epsilon$$ as long as $n>n_0$
We start from:
$$\left|\left(\frac{ni}{2^n}\right)\right|=|i||\frac{n}{2^n}|=|\frac{n}{2^n}|$$
On the other hand, what I did was to prove these inequalities by induction:
$$2^{n+1}<3^n$$ if $n\geq2$ $$n*2^n<3^n$$ if $n\geq2$
to arrive that $$\forall{n\geq2}:0<\frac{n}{2^n}<(\frac{3}{4})^n$$
and, knowing that $$\lim _{n\to\infty}\left((\frac{3}{4})^n\right)=0$$ we can use the Sandwich Lemma to prove that $$\lim_{n\to\infty}\left(\frac{n}{2^n}\right)=0$$
But that's not what I'm interested in, because I'm not using the definition. Even if I go back to $$\left|\left(\frac{ni}{2^n}\right)\right|=|i||\frac{n}{2^n}|=|\frac{n}{2^n}|=\frac{n}{2^n}<(\frac{3}{4})^n<\epsilon$$ so how much is $n$ for this inequality to hold? What value of $n$ should I take for the definition to be fulfilled?
You have a simpler path since in fact $n^\alpha=o(2^n)$ for all powers of $n$.
So you can prove by induction $2^n\ge n^2$ for $n\ge 4$
Induction step goes as $2^{n+1}-(n+1)^2=2\times\underbrace{(2^n-n^2)}_\text{positive by induction}+\underbrace{(n^2-2n-1)}_\text{positive for n>3}\ge 0$
As a result your expression $\left|\dfrac{ni}{2^n}\right|\le \dfrac 1n$ and this is trivial to find a delta-epsilon.