Using power series of e^x, show that if N= 1+1/2+1/3.1/n then e^N>n+1 hence prove series diverges?
help!
Answers:
According to the properties of the exponencial function, for every real x we have e^x >= 1+ x, with equality if and only if x = 0. Therefore, for every naturaln, it follows that:
e^1 > (1 +1/1)
e^2 > (1 +1/2)
.
.
e^n > (1 +1/n)
Since all terms involved in such inequalities are positive, if we multiply such inequalties and apply the properties of the exponencial function, we get
e^(1 + 1/2 ..+ 1/n) = e^N > (1+1/1) * (1 +1/2)*..(1 +1/n) On the right hand side, we have n terms and n-1 of them are greater than (1 + 1/n). Therefore, e^N > n * (1+ 1/n) = n +1, proving your inequalty. Taking natural logarithms, it follows that N > ln( n+1) for n =1,2,3. Snce ln(n+1) -> oo when n -> oo, it follows N -> oo when n -> oo, so proving the harmonic series diverges.
N= 1+1/2+1/3.1/n
since
1+1/2+1/3.1/n >
{1+1/2 +1/4+1/8 +1/6+1/32....n} terms
Since RH is in G.P. whose first term is 1 and common ratio is (1/2)
Sn=1{1-0.5^n)/(1-0.5)
=2{1-(1/2)^n}=2-(1/2)^(n-1)
1+1/2+1/3.1/n>2-(1/2)^(n-1)
Therefore
e^(1+1/2+1/3.1/n)>e^{2-(1/2).
when n tends to infinity then RH=e^2
Therefore it is divergent series
e^N>
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Answers:
According to the properties of the exponencial function, for every real x we have e^x >= 1+ x, with equality if and only if x = 0. Therefore, for every naturaln, it follows that:
e^1 > (1 +1/1)
e^2 > (1 +1/2)
.
.
e^n > (1 +1/n)
Since all terms involved in such inequalities are positive, if we multiply such inequalties and apply the properties of the exponencial function, we get
e^(1 + 1/2 ..+ 1/n) = e^N > (1+1/1) * (1 +1/2)*..(1 +1/n) On the right hand side, we have n terms and n-1 of them are greater than (1 + 1/n). Therefore, e^N > n * (1+ 1/n) = n +1, proving your inequalty. Taking natural logarithms, it follows that N > ln( n+1) for n =1,2,3. Snce ln(n+1) -> oo when n -> oo, it follows N -> oo when n -> oo, so proving the harmonic series diverges.
N= 1+1/2+1/3.1/n
since
1+1/2+1/3.1/n >
{1+1/2 +1/4+1/8 +1/6+1/32....n} terms
Since RH is in G.P. whose first term is 1 and common ratio is (1/2)
Sn=1{1-0.5^n)/(1-0.5)
=2{1-(1/2)^n}=2-(1/2)^(n-1)
1+1/2+1/3.1/n>2-(1/2)^(n-1)
Therefore
e^(1+1/2+1/3.1/n)>e^{2-(1/2).
when n tends to infinity then RH=e^2
Therefore it is divergent series
e^N>
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