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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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    $\begingroup$ Your answer is incorrect because the question is asking whether the statement is true for every sequence $\{a_n\}$. You have only proved it in the special case when $a_n=0$ for every $n$. $\endgroup$ – Prism Dec 23 '14 at 1:47
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    $\begingroup$ "Does there exist…" part is about existence of the series whose partial sums is $\{a_n\}$. Let's look at the beginning of the question. It says "Let $\{a_n\}_{n=1}^{\infty}$ be an infinite sequence of real numbers…" It is an arbitrary sequence of real numbers! $\endgroup$ – Prism Dec 23 '14 at 1:50
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    $\begingroup$ It is an existence question given an arbitrary sequence. So it is really a universal question: does a certain property hold for any sequence. $\endgroup$ – Ian Dec 23 '14 at 1:50
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    $\begingroup$ @Ian : I'd have phrased it thus: "does a certain property hold for every sequence?". The standard way of using the word "any" would admit the following way of misunderstanding your phrasing: "does a certain property hold for any sequence" could mean "Is there any sequence for which a certain property holds?". But that cannot be what was meant. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 23 '14 at 2:03
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    $\begingroup$ @MichaelHardy You're right, the word "any" is frustrating for this exact reason. $\endgroup$ – Ian Dec 23 '14 at 2:19
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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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  • $\begingroup$ Please elaborate $\endgroup$ – Al Jebr Dec 23 '14 at 1:57
  • $\begingroup$ @JohannFranklin : I'm not sure there's much else to say. If the partial sum $b_0+\cdots+b_{100}$ is $a_{100}$, and we want the partial sum $b_0+\cdots+b_{100}+b_{101}$ to be $a_{101}$, then we have to add $b_{101}$ to the sum we've already got, which is $a_{100}$, to get $a_{101}$. So the question is: what do we have to add to $a_{100}$ to get $a_{101}$? The answer is $a_{101}-a_{100}$, so we need $b_{101}$ to be equal to that. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 23 '14 at 2:00
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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

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

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