sum of infinite series: 1+3/4+7/16+15/64+...

[wallflower]

​What is the sum of infinite series 1+3/4+7/16+15/64+.........

 

[Deleted member 4993]

​What is the sum of infinite series 1+3/4+7/16+15/64+.........

What is the n th term of this series?

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[pka]

​What is the sum of infinite series 1+3/4+7/16+15/64+.........

\(\displaystyle \displaystyle\sum\limits_{n = 1}^\infty {\frac{{{2^n} - 1}}{{{4^{n-1} }}}} = ?\)
You post the answer!

 

[HallsofIvy]

\(\displaystyle \displaystyle\sum\limits_{n = 1}^\infty {\frac{{{2^n} - 1}}{{{4^{n-1} }}}} = ?\)
You post the answer!

Which is, of course, \(\displaystyle \displaystyle\sum\limits_{n = 1}^\infty \frac{2^n}{4^n} - \sum\limits_{n = 1}^\infty \frac{1}{4^{n-1}}\)
\(\displaystyle = \sum\limits_{n = 1}^\infty \frac{1}{2^n}- 4\sum\limits_{n = 1}^\infty \frac{1}{4^n}\).
Both are geometric series.

 

[pka]

Which is, of course, \(\displaystyle \displaystyle\sum\limits_{n = 1}^\infty \frac{2^n}{4^n} - \sum\limits_{n = 1}^\infty \frac{1}{4^{n-1}}\)
\(\displaystyle = \sum\limits_{n = 1}^\infty \frac{1}{2^n}- 4\sum\limits_{n = 1}^\infty \frac{1}{4^n}\).
Both are geometric series.

Spoil-sport.

 

[Otis]

Spoil-sport.

You, too, because you answered Subhotosh's question to the student.

 

[lookagain]

\(\displaystyle \displaystyle\sum\limits_{n = 1}^\infty {\frac{{{2^n} - 1}}{{{4^{n-1} }}}} = ?\)

\(\displaystyle = \ \displaystyle\sum\limits_{n = 1}^\infty \frac{2^n}{4^n} - \sum\limits_{n = 1}^\infty \frac{1}{4^{n-1}}\)
\(\displaystyle = \sum\limits_{n = 1}^\infty \frac{1}{2^n}- 4\sum\limits_{n = 1}^\infty \frac{1}{4^n}\).

The expression to the left of the subtraction sign is different:

\(\displaystyle = \ \displaystyle\sum\limits_{n = 1}^\infty \frac{2^n}{4^{n-1}} - \sum\limits_{n = 1}^\infty \frac{1}{4^{n-1}}\)

\(\displaystyle = \ 4\displaystyle\sum\limits_{n = 1}^\infty \frac{2^n}{4^n} - 4\sum\limits_{n = 1}^\infty \frac{1}{4^{n}}\)

\(\displaystyle = 4\sum\limits_{n = 1}^\infty \frac{1}{2^n}- 4\sum\limits_{n = 1}^\infty \frac{1}{4^n}\)