Assume that an→ℓ is a convergent sequence of complex numbers and {λn} is a sequence of positive real numbers such that ∞∑k=0λk=∞
Then, show that,
limn→∞1n∑k=0λkn∑k=0λkak=ℓ=limn→∞an
(Note that : This is more general than the special case where, λn=1)
Answer
Let ε>0 and Nsuch that |ak−l|≤ε for all k>N
Then, for n>N we have,
|n∑k=0λkakn∑k=0λk−l|=|n∑k=0λk(ak−l)n∑k=0λk|=|N∑k=0λk(ak−l)+n∑k=Nλk(ak−l)n∑k=0λk|≤Mn∑k=0λk+n∑k=Nλk|ak−l|⏟≤εn∑k=0λk≤Mn∑k=0λk+ε→0
since N∑k=0λk→∞.
Where M=|N∑k=0λk(ak−l)|
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