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JH
Numerade Educator

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Problem 85 Medium Difficulty

If $ \sum a_n $ is convergent and $ \sum b_n $ is divergent, show that the series $ \sum \left( a_n + b_n \right) $ is divergent. [Hint: Argue by contradiction.]

Answer

$\sum b_{n}$ is given to be divergent.

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Video Transcript

were given that this sum of am convergence both the sum of being diverges and we'LL use this to show that the Siri's of A and plus beyond his diversion and the hit is that we should argue by contradiction. So let's follow the hidden here. So let's go ahead and suppose that this sum of a M plus B end conversions then if we take this summon, subtract this sum. This will also converge because we're just subtracting two real numbers here. If this some conversions, that means that the sum of a and lesbians a real number. Similarly, we're already assuming that this son commercials that means the sums of real number. So when I subtract, this is also a real number. So here, let me not say conversions. I'll just say this is a real number. However, this Siri's here. We can also rewrite this is and plus being minus and and then here because we're just dealing with real numbers on the inside. Well, you just go ahead and cancel those ends. We have the sum of bien. So on one hand, this expression here, the difference of the two sums is a real number on the other hand, this sum is also equal to the sum of the end. But we're told that this diverges that was the assumptions. So it is impossible for the sum of the end to be a real number and diverge. It's one or the other. If it's equal to a real number, that would mean converges. Excuse me. Therefore, we've arrived at the contradiction, and that's our final answer.