How do you determine if #Sigma (7*5^n)/6^n# from #n=[0,oo)# converge and find the sums when they exist?

Answer 1

#sum_(n=0)^oo (7*5^n)/6^n = 42#

Write the series as:

#sum_(n=0)^oo (7*5^n)/6^n = 7 sum_(n=0)^oo (5/6)^n#
so this is a geometric series of ratio #r=5/6#.
As #abs(r) < 1# the series is convergent and its sum is:
#7 sum_(n=0)^oo (5/6)^n =7/(1-5/6) = 7/(1/6) = 42#
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Answer 2

To determine the convergence of the series ( \sum_{n=0}^{\infty} \frac{7 \cdot 5^n}{6^n} ), we can use the ratio test.

  1. Apply the ratio test: ( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} ) ( = \lim_{n \to \infty} \frac{7 \cdot 5^{n+1}/6^{n+1}}{7 \cdot 5^n/6^n} ) ( = \lim_{n \to \infty} \frac{5^{n+1}}{6^{n+1}} \cdot \frac{6^n}{5^n} ) ( = \lim_{n \to \infty} \frac{5}{6} ) ( = \frac{5}{6} )

  2. Since the limit is less than 1, the series converges.

To find the sum: ( S = \frac{a}{1 - r} ) ( S = \frac{7 \cdot \frac{5^0}{6^0}}{1 - \frac{5}{6}} ) ( S = \frac{7}{1 - \frac{5}{6}} ) ( S = \frac{7}{\frac{1}{6}} ) ( S = 42 )

Therefore, the series converges, and its sum is 42.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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