Using the first 10 terms graph the sequence, and use the graph to discuss the convergence or divergence of the sequence an=3-1/2^n?

1 Answer
Apr 9, 2018

Converges to #3# when examining the graph.

Explanation:

So, we have the sequence #a_n=3-(1/2)^n.#

Assuming #n>=1#, we need to calculate the first ten values.

#a_1=3-1/2=5/2#

#a_2=3-(1/2)^2=3-1/4=11/4#

#a_3=3-(1/2)^3=3-1/8=23/8#

#a_4=3-(1/2)^4=3-1/16=47/16#

#a_5=3-(1/2)^5=3-1/32=96/32#

#a_6=3-(1/2)^6=3-1/64=191/64#

#a_7=3-(1/2)^7=3-1/128=383/128#

#a_8=3-(1/2)^8=3-1/256=767/256#

#a_9=3-(1/2)^9=3-1/512=1535/512#

#a_10=3-(1/2)^10=3-1/1024=3071/1024#

Now, we plot the following points:

#(1, 5/2), (2, 11/4), (3, 23/8), (4, 47/16), (5, 96/32), (6, 191/64), (7, 383/128), (8,767/256), (9, 1535/512), (10, 3071/1024)#

But we do not connect these points, as a sequence only returns outputs for integers. Connecting points would imply the sequence exists everywhere between the plotted ploints, which is absolutely false:

enter image source here

This program connected the points together with straight lines to make visualization easier, but do not actually connect the points together with lines when making your plot. The sequence only exists at the plotted points.

A sequence #a_n# converges if, for increasing values of #n, a_n# gets closer to some finite value, IE, #lim_(n->oo)a_n ne +-oo#.

From the plot, we can see this sequence converges. It approaches #y=3.0#, with the growth slowing down as we get closer and closer.