What is the sum of the infinite geometric series #sum_(n=1)^oo2^n/5^(n-1)# ?

Answer 1
This series can be rewritten as #sum_{n=1}^\infty 2(2/5)^{n-1}# and is thus a geometric series with first term 2 and common ratio #2/5#. Since the common ratio is less than 1 (in absolute value), the series is convergent and its sum is given by #"first term"/(1-"common ratio")=2/(1-2/5)=10/3#. See Geometric Series
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Answer 2

The sum of the infinite geometric series ( \sum_{n=1}^{\infty} \frac{2^n}{5^{n-1}} ) can be calculated using the formula for the sum of an infinite geometric series, which is ( \frac{a}{1 - r} ), where ( a ) is the first term and ( r ) is the common ratio. In this series, the first term ( a ) is ( \frac{2}{5} ) and the common ratio ( r ) is ( \frac{2}{5} ). Therefore,

[ \sum_{n=1}^{\infty} \frac{2^n}{5^{n-1}} = \frac{\frac{2}{5}}{1 - \frac{2}{5}} ]

[ = \frac{\frac{2}{5}}{\frac{3}{5}} ]

[ = \frac{2}{3} ]

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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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