Question

sum of infinite over E n=1 6(1/3)^n-1

Answer 1

Answer:

$\displaystyle \large{\sum_{n=1}^\infty 6\left(\dfrac{1}{3}\right)^{n-1} = \dfrac{6}{1-\dfrac{1}{3}}}$

Step-by-step explanation:

The given summation is infinite geometric series as it’s shown an exponent which geometric series/sequence has.

Let’s review partial geometric series formula:

$\displaystyle \large{S_n=\dfrac{a_1(1-r^n)}{1-r}}$

When we input the limit as n approaches infinity:

$\displaystyle \large{\lim_{n\to \infty} S_n = \lim_{n\to \infty} \dfrac{a_1(1-r^n)}{1-r}}$

Consider each intervals.

( 1 ) When |r| > 1

Since |r| is greater than 1, it’s defined to approach infinity. Since it approaches infinity then we can say that the series diverges to infinity.

( 2 ) When r = 1

For r = 1, the equation for partial sum is S = na where a is first term and n is number of first term. For r = 1, the sum will continue to grow and will eventually approaches infinity. Henceforth, the series diverges.

( 3 ) When r ≤ -1

In this interval, the series oscillates which is defined to be divergent. Hence, the series diverges.

( 4 ) When |r| < 1

In this interval, $\displaystyle \large{r^n}$ will approach 0 as accorded to exponential graph.

$\displaystyle \large{\lim_{n\to \infty} S_n = \lim_{n\to \infty} \dfrac{a_1(1-0)}{1-r}}$

Hence, for |r| < 1, the geometric series will converge to:

$\displaystyle \large{\lim_{n\to \infty} S_n = \dfrac{a_1}{1-r}}$

We can also write in summation notation form as:

$\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}$

Therefore, from the given problem, it’s fortunately that the series is already in the same pattern as formula. Therefore, we simply apply the formula in.

$\displaystyle \large{\sum_{n=1}^\infty 6\left(\dfrac{1}{3}\right)^{n-1} = \dfrac{6}{1-\dfrac{1}{3}}}$

Therefore, the answer is third choice! Please let me know if you have any questions!