I need to evaluate the following limit :

Suppose that $\displaystyle\lim_{n \to \infty} a_n = a$ then find the Limit :

$\displaystyle\lim_{n \to \infty} \left(\dfrac{a_n}{1} + \dfrac{a_{n-1}}{2} + \dfrac{a_{n-2}}{2^2} + \ldots + \dfrac{a_1}{2^{n-1}}\right)$

Since term in denominators forms a Geometric sequence,I am trying to solve this by using Sandwich Theorem to get some insights but I haven't achieved any success .

Any idea how should I proceed towards the solution ?


We have that

$$\sum_{k=1}^n \frac{a_k}{2^{n-k}}=\frac1{2^n}\sum_{k=1}^n a_k2^k$$

then we can use Stolz-Cesaro theorem.


@gimusi already gave a one-liner proof, so let me introduce another line of argument. In this answer, we use the following theorem whose proof is quite standard and no more demanding than proving OP's specific case.

Theorem. (Dominated Convergence Theorem for Series) Let $(c_{k,n})_{k,n\geq 1}$ and $(M_k)_{k\geq 1}$ be real numbers such that

  1. $|c_{k,n}| \leq M_k$ for all $n, k \geq 1$,
  2. $\lim_{n\to\infty} c_{k,n} $ converges for each $k \geq 1$,
  3. $\sum_{k=1}^{\infty} M_k < \infty$.


$$ \lim_{n\to\infty} \sum_{k=1}^{\infty} c_{k,n} = \sum_{k=1}^{\infty} \lim_{n\to\infty} c_{k,n}. $$

In our case, the sequence $(a_n)_{n\geq 1}$ is bounded by, say $M > 0$, and so, we can set

$$ c_{k,n} = \frac{a_{n-k+1}}{2^{k-1}} \qquad \text{and} \qquad M_k = \frac{M}{2^{k-1}}. $$

Then the limit is $\sum_{k=1}^{\infty} \frac{a}{2^{k-1}} = 2a$. So it remains to prove the theorem.

Proof of Theorem. Write $\alpha_k = \lim_{n\to\infty} c_{k,n}$ and note that $|\alpha_k| \leq M_k$ for each $k\geq 1$, and so, both $\sum_{k=1}^{\infty} c_{k,n}$ and $\sum_{k=1}^{\infty} \alpha_k$ converge absolutely. Then for each fixed $N \geq 1$,

$$ \left|\sum_{k=1}^{\infty} c_{k,n} - \sum_{k=1}^{\infty} \alpha_k \right| \leq \sum_{k=1}^{\infty} |c_{k,n} - \alpha_k| \leq \sum_{k=1}^{N} |c_{k,n} - \alpha_k| + \sum_{k=N+1}^{\infty} 2M_k. $$

So it follows that

$$ \limsup_{n\to\infty} \left|\sum_{k=1}^{\infty} c_{k,n} - \sum_{k=1}^{\infty} \alpha_k \right| \leq \sum_{k=N+1}^{\infty} 2M_k. $$

But since the left-hand side is independent of $N$, letting $N \to \infty$ shows that the left-hand side is $0$, and so, the desired claim follows. $\square$


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