Determine if the sequence is convergent or divergent. If it converges, find the limit. (ln(1+2e^n))/n

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lemjay | High School Teacher | (Level 3) Senior Educator

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The given sequence is:

`a_n=ln(1+2e^n)/n`

Take note that a sequence is convergent if the terms approaches a real number.

Let's list down the some of the terms of this sequence.

`a_1 = ln(1+2e^1)/1=1.8620`

`a_2=ln(1+2e^2)/2=1.3793`

`a_3=ln(1+2e^3)/3=1.2392`

`a_4=ln(1+2e^4)/4=1.1756`

`a_5=ln(1+2e^5)/5=1.1393`

`a_6=ln(1+2e^6)/6=1.1157`

`a_7=ln(1+2e^7)/7=1.0991`

So the first few terms of the sequence are:

`{a_n} = {1.8620, 1.3793,1.2392,1.1756,1.1393,1.1157,1.0991, ...}`

Notice that as n increases, the values of `a_n` decrease and seem to approach 1. To verify this, let's determine its limiting value as n approaches infinity.

`lim_(n->oo) a_n = lim_(n->oo) ln(1+2e^n)/n`

To take its limit, apply L'Hospital's Rule. So take the derivative of the numerator and denominator.

`lim_(n->oo) ((ln(1+2e^n))')/((n)')=lim_(n->oo) (1/(1+2e^n)*2e^n)/1 = lim_(n->oo) (2e^n)/(1+2e^n)`

Then, plug-in `n=oo` .

`lim_(n->oo) (2e^n)/(1+2e^n) = (2e^oo)/(1+2e^oo)=oo/oo`

Notice that the result is indeterminate. So to  find its limit, we have to take the derivative of the numerator and denominator again.

`lim_(n->oo) ((2e^n)')/((1+2e^n)') = lim_(n->oo) (2e^n)/(2e^n) = lim_(n->oo) 1`

Take note that if we take the limit of the constant, the result is the constant itself.

`lim_(n->oo)1= 1`

Therefore, the sequence converges and its limit is 1.

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