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March 2nd, 2017, 02:54 AM   #1
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Hi...I'm a new member here...I would be glad if anyone could see the following is cor



Shenaya Perera
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March 2nd, 2017, 05:17 AM   #2
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The left side of that is rather blurry and hard to read but I believe you have


You then divide both numerator and denominator by , dividing each term in both numerator and denominator by that. Yes, since the numerator is a finite sum, you can do that. Each term in the numerator is
getting
.

Now, again because the numerator is a finite sum with 2017 terms, no matter what x is, you can take the limit in each term as x goes to infinity. Of course, each fraction, with x in the denominator, goes to 0 leaving the sum of 2017 "1"s in the numerator while the denominator goes to 1. The result is 2017.
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March 2nd, 2017, 05:24 AM   #3
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Quote:
Originally Posted by Country Boy View Post
The left side of that is rather blurry and hard to read but I believe you have


You then divide both numerator and denominator by , dividing each term in both numerator and denominator by that. Yes, since the numerator is a finite sum, you can do that. Each term in the numerator is
getting
.

Now, again because the numerator is a finite sum with 2017 terms, no matter what x is, you can take the limit in each term as x goes to infinity. Of course, each fraction, with x in the denominator, goes to 0 leaving the sum of 2017 "1"s in the numerator while the denominator goes to 1. The result is 2017.
Sorry for the inconveniece caused and thanks for the answer and your explanation,now I'm sure I've done this in the correct way!

Shenaya Perera
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