Let $\{a_n\}_{n=1}^\infty$ be an infinite sequence. Does there exist an infinite series whose partial sums is $\{a_n\}_{n=1}^\infty$?

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Definition: By an Infinite Sequence of real numbers, we shall mean any real valued function whose domain is the set of all positive integers.

Definition: By an Infinite Series of real numbers, we shall mean an ordered pair of infinite sequences of real numbers $$(\{a_n\}_{n=1}^\infty, \{s_n\}_{n=1}^\infty)$$ such that $$s_1=a_1, \space s_2=a_1+a_2, \space s_3=a_1+a_2+a_3$$ and in general $$s_k= a_1+a_2+ \dotsb +a_k$$ The sequence $\{a_n\}_{n=1}^\infty$ is called the sequence of terms of the infinite series. The sequence $\{s_n\}_{n=1}^\infty$ is called the sequence of partial sums of the infinite series.

Exam Question:

Let $\{a_n\}_{n=1}^\infty$ be an infinite sequence of real numbers. Does there exist an infinite series of real numbers whose sequence of partial sums is $\{a_n\}_{n=1}^\infty$? Why or why not?

My answer is Yes and here is why. Consider the constant sequence $\{0\}_{n=1}^\infty$. This sequence has the value $0$ for all $n \in \mathbb Z^+$. Now, if we consider the infinite series of this sequence, then the sequence of partial sums has the value $0$ for all $n \in \mathbb Z^+$. So, in this case $$(\{a_n\}_{n=1}^\infty, \{s_n\}_{n=1}^\infty) = (\{0\}_{n=1}^\infty, \{0\}_{n=1}^\infty) = (0,0)$$

Therefore, $\{a_n\}_{n=1}^\infty = 0 = \{s_n\}_{n=1}^\infty$ for all $n \in \mathbb Z^+$.

However, on turning in my exam, my professor stated this was incorrect. Why is this so?

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Certainly: $$ \text{Let }b_0 = a_0\text{ and }b_{n+1} = a_{n+1}-a_n. $$ Then \begin{align} b_0 & = a_0 & = a_0 \\ b_0 + b_1 & = a_0 + (a_1-a_0) & = a_1 \\ b_0 + b_1 + b_2 & = a_0 + (a_1-a_0)+(a_2-a_1) & = a_2 \\ b_0 + b_1 + b_2 + b_3 & = a_0 + (a_1-a_0)+(a_2-a_1)+(a_3-a_2) & = a_3 \\ & {}\,\,\vdots & \vdots\phantom{a_n} \end{align}

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Because it only works when $a_n=0$ for all $n$. Let $s_0=a_0$, and $s_n=a_n-a_{n-1}$. Now

$$\sum_{n=0}^ks_n=a_0+(a_1-a_0)+\dots +(a_k-a_{k-1})=a_k.$$