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How do I compute the infinite sum $$\sum_{k=1}^{\infty} (-\cfrac13)^{k-1}$$

I notice that values oscillate between positive and negative numbers depending on whether $k$ is even or odd, but on a graph this will still converge quickly to the x-axis because $3^{k-1}\ge1^{k-1}$, which suggests the total sum $P$ does exist.

When I graphed it I notice after $k=6$ the functions $\Delta y$ is so little it seems insignificant to add more terms. Can I just get away with adding the values up to $x=6$? Doing this I get 182/243, roughly $0.748$. I then imagine this suggests $P = 3/4$, which happens to be one of the possible answers in my homework.

Is there a better approach to this or am I good here?

Lex_i
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  • There is a closed expression for those type of sums: $$\frac{1}{1-r},$$ where $r$ is deduced from the sum: $$\sum_{k=0}^\infty r^k.$$ In order to apply the formula above you need to convert your sum into that expression. – Dog_69 Jan 13 '20 at 22:31

2 Answers2

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Hint: calculate the finite sum $\sum_{k=1}^{N}\left(-\frac{1}{3}\right)^{k-1}$ (geometric progression) and then let $N\to\infty$.

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I'll attempt to address some mistakes you've made.

...on a graph this will still converge quickly to the x-axis because $3^{k−1}≥1^{k−1}$, which suggests the total sum P does exist.

If I'm understanding you correctly, it is not enough to infer that $P$ exists from just knowing that the denominators are successively larger, i.e. $3^{k−1}≥1^{k−1}$. There are counterexamples to this, e.g., the harmonic series, $\displaystyle \frac11+\frac12+\frac13+\ldots=\infty$, which doesn't converge. In order to show that $P$ exists, we would need to know that successive terms are sufficiently (e.g. $3$ times) smaller.

When I graphed it I notice after $k=6$ the functions $Δy$ is so little it seems insignificant to add more terms. Can I just get away with adding the values up to $x=6$?

This is called computing the partial sum of an infinite series. If the series is convergent then a partial sum can give you an approximate ballpark for the value of the whole series and this can allow you to conjecture a value. However, this is not a proof that the series equals that value. In order to definitively prove the value of the series, you would need one of the methods shown in question 29023. Also, from a practical point of view, in this case, it is far easier to compute the value at infinity than at $x=6$, so that is the clear choice in an exam setting.

Jam
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