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Proof of Limit Comparison Test
As 0 < c <∞ we can find out two positive and finite numbers, m and M, like m < c < M .
Now, as
we know that for large enough n the quotient an/bn should be close to c and thus there must be a positive integer N like if n > N we as well have,
m < an / bn < M
Multiplying by bn provides
Mbn < an < Mbn
provided n > N .
Here now, if ∑bn diverges then thus does ∑mbn and so as mbn < an for all adequately large n by the Comparison Test ∑an as well diverges.
Similarly, if ∑bn converges then so does ∑Mbn and as an < Mbn for all sufficiently large n by the Comparison Test ∑an as converges.
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report on shares and dividend using newspaper
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