There is the following sum: $1-\sum\limits_{k=1}^{n}\frac{2^{k+1}}{3^k}\alpha=1-\alpha(1-(\frac{2}{3})^n)$ where $\alpha\in(0,1]$

I do not understand the following equality. I thought it could be derived using geometric series sum:



Why is the equality ,I derived, different from the initial one presented?

Thanks in advance!

  • $\begingroup$ @Winther The answer made me spot the mistake when copiying from the book. In the original rext it is written ($2^{k-1}$). It was not with the any intention of invalidating the answer. $\endgroup$ – Pedro Gomes Dec 1 '18 at 14:41

I believe you have a typo in the question statement.

$$1-\sum\limits_{k=1}^{n}\frac{2^{k\color{red}-1}}{3^k}\alpha=1-2^{-1}\alpha\sum\limits_{k=1}^{n}\frac{2^{k}}{3^k}=1-2^{-1}\left( \frac23\right)\alpha(\frac{1-(\frac{2}{3})^n}{\frac{1}{3}})=1-\alpha(1-(\frac{2}{3})^n)$$

and you forgot the first term in the geometric sum.

  • $\begingroup$ Yes you are right about the typo. $\endgroup$ – Pedro Gomes Dec 1 '18 at 14:42

Your mistake is in the statement $$ 1 - \sum_{k=1}^{n} \frac{2^{k+1}}{3^k} = 1 - 2^{-1} \alpha \sum_{k=1}^{n} \frac{2^k}{3^k}. $$ The correct expression is $$ 1 - \sum_{k=1}^{n} \frac{2^{k+1}}{3^k} = 1 - 2 \alpha \sum_{k=1}^{n} \frac{2^k}{3^k}, $$ since $2 \times 2^k = 2^{k+1}$.

  • $\begingroup$ Typo was fixed. Thanks for pointing that out! $\endgroup$ – Pedro Gomes Dec 1 '18 at 14:36

You need to consider the first term of a geometric series.


$$u_1 = \frac{2}{3} \quad r = \frac{2}{3}$$


$$\sum_{k = 1}^{n}\bigg(\frac{2}{3}\bigg)^k = \sum_{k = 1}^{n}\frac{2}{3}\cdot\bigg(\frac{2}{3}\bigg)^{k-1} = \frac{\frac{2}{3}\big(1-\big(\frac{2}{3}\big)^n\big)}{1-\frac{2}{3}} = 2\bigg(1-\bigg(\frac{2}{3}\bigg)^n\bigg)$$

Your error was that you took $u_1 = 1$, leading to an incorrect result.


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